Answer

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Answer

The equilibrium constant $(K_c)$ for the give reaction is:

$ \begin{aligned} K_c & =\frac{[SO_3]^{2}}{[SO_2]^{2}[O_2]} \\ & =\frac{(1.90)^{2} M^{2}}{(0.60)^{2}(0.821) M^{3}} \\ & =12.239 M^{-1} \text{ (approximately) } \end{aligned} $

Hence, $K_c$ for the equilibrium is $12.239 M^{-1}$.

Answer

Partial pressure of I atoms,

$p_1=\frac{40}{100} \times p _{\text{total }}$

$=\frac{40}{100} \times 10^{5}$

$=4 \times 10^{4} Pa$

Partial pressure of $I_2$ molecules,

$ \begin{aligned} & p _{I_2}=\frac{60}{100} \times p _{\text{total }} \\ & =\frac{60}{100} \times 10^{5} \\ & =6 \times 10^{4} Pa \end{aligned} $

Now, for the given reaction,

$ \begin{aligned} K_p & =\frac{(p I)^{2}}{p _{I_2}} \\ & =\frac{(4 \times 10^{4})^{2} Pa^{2}}{6 \times 10^{4} Pa} \\ & =2.67 \times 10^{4} Pa \end{aligned} $

Answer

(i) $\quad K_c=\frac{[NO _{(g)}]^{2}[Cl _{2(g)}]}{[NOCl _{(g)}]^{2}}$

(ii) $\quad K_c=\frac{[CuO _{(s)}]^{2}[NO _{2(g)}]^{4}[O _{2(g)}]}{[Cu(NO_3) _{2(s)}]^{2}}$

$ =[NO _{2(g)}]^{4}[O _{2(g)}] $

(iii) $K_c=\frac{[CH_3 COOH _{(a q)}][C_2 H_5 OH _{(a q)}]}{[CH_3 COOC_2 H _{5(a q)}][H_2 O _{(l)}]}=\frac{[CH_3 COOH _{(a q)}][C_2 H_5 OH _{(a q)}]}{[CH_3 COOC_2 H _{5(a q)}]}$

(iv) $K_c=\frac{[Fe(OH) _{3(s)}]}{[Fe _{(a q)}^{3+}][OH _{(a q)}^{-}]^{3}}$

$ =\frac{1}{[Fe _{(a q)}^{3+}][OH _{(a q)}^{-}]^{3}} $

(v) $\quad K_c=\frac{[IF_5]^{2}}{[I _{2(s)}][F_2]^{5}}$

$=\frac{[IF_5]^{2}}{[F_2]^{5}}$

Answer

The relation between $K_p$ and $K_c$ is given as:

$K_p=K_c(RT)^{\Delta n}$

(a) Here,

$\Delta n=3-2=1$

$R=0.0831$ barLmol $^{-1} K^{-1}$

$T=500 K$

$K_p=1.8 \times 10^{-2}$

Now,

$K_p=K_c(R T)^{\Delta n}$

$\Rightarrow 1.8 \times 10^{-2}=K_c(0.0831 \times 500)^{1}$

$\Rightarrow K_c=\frac{1.8 \times 10^{-2}}{0.0831 \times 500}$

$ =4.33 \times 10^{-4} \text{ (approximately) } $

(b) Here,

$\Delta n=2$ - $1=1$

$R=0.0831$ barLmol $^{-1} K^{-1}$

$T=1073 K$

$K_p=167$

Now,

$K_p=K_c(R T)^{\Delta n}$

$\Rightarrow 167=K_c(0.0831 \times 1073)^{\Delta n}$

$\Rightarrow K_c=\frac{167}{0.0831 \times 1073}$

$=1.87$ (approximately)

Answer

It is given that $K_C$ for the forward reaction is $6.3 \times 10^{14}$.

Then, $K_C$ for the reverse reaction will be,

$ \begin{aligned} K_C^{\prime} & =\frac{1}{K_C} \\ & =\frac{1}{6.3 \times 10^{14}} \\ & =1.59 \times 10^{-15} \end{aligned} $

Answer

For a pure substance (both solids and liquids),

$ \begin{aligned} {[\text{ Pure substance }] } & =\frac{\text{ Number of moles }}{\text{ Volume }} \\ & =\frac{\text{ Mass } / \text{ molecular mass }}{\text{ Volume }} \\ & =\frac{\text{ Mass }}{\text{ Volume } \times \text{ Molecular mass }} \\ & =\frac{\text{ Density }}{\text{ Molecular mass }} \end{aligned} $

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Answer

Let the concentration of $N_2 O$ at equilibrium be $x$.

The given reaction is:

Therefore, at equilibrium, in the $10 L$ vessel:

$[N_2]=\frac{0.482-x}{10},[O_2]=\frac{0.933-x / 2}{10},[N_2 O]=\frac{x}{10}$

The value of equilibrium constant i.e., $K_C=2.0 \times 10^{9 a+37}$ is very small. Therefore, the amount of $N_2$ and $O_2$ reacted is also very small. Thus, $x$ can be neglected from the expressions of molar concentrations of $N_2$ and $O_2$.

Then,

$[N_2]=\frac{0.482}{10}=0.0482 mol L^{-1}$ and $[O_2]=\frac{0.933}{10}=0.0933 mol L^{-1}$

Now,

$ \begin{aligned} & K_C=\frac{[N_2 O _{(g)}]^{2}}{[N _{2(g)}]^{2}[O _{2(g)}]} \\ & \Rightarrow 2.0 \times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)} \\ & \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37} \times(0.0482)^{2} \times(0.0933) \\ & \Rightarrow x^{2}=43.35 \times 10^{-40} \\ & \Rightarrow x=6.6 \times 10^{-20} \\ & {[N_2 O]=\frac{x}{10}=\frac{6.6 \times 10^{-20}}{10}} \\ & =6.6 \times 10^{-21} \end{aligned} $

Answer

The given reaction is:

$ 2 NO _{(g)}+Br _{2(g)} \longleftrightarrow 2 NOBr _{(g)} $

$2 mol \quad 1 mol \quad 2 mol$

Now, 2 mol of $NOBr$ are formed from $2 mol$ of $NO$. Therefore, $0.0518 mol$ of $NOBr$ are formed from $0.0518 mol$ of $NO$.

Again, 2 mol of $NOBr$ are formed from $1 mol$ of $Br$.

Therefore, $0.0518 mol$ of $NOBr$ are formed from $\frac{0.0518}{2} mol$ of $Br$, or $0.0259 mol$ of NO.

The amount of $NO$ and $Br$ present initially is as follows:

$[NO]=0.087 mol[Br_2]=0.0437 mol$

Therefore, the amount of NO present at equilibrium is:

$[N O]=0.087$ - 0.0518

$=0.0352 mol$

And, the amount of $Br$ present at equilibrium is:

$[Br_2]=0.0437$ - 0.0259

$=0.0178 mol$

Answer

For the given reaction,

$\Delta n=2$ - 3 = - 1

$T=450 K$

$R=0.0831$ bar $L$ bar $K^{- 1 }$

$K_p=2.0 \times 10^{10} bar^{-1}$

We know that,

$ \begin{aligned} & K_P=K_C(R T) \Delta n \\ & \begin{aligned} & \Rightarrow 2.0 \times 10^{10} bar^{-1}=K_C(0.0831 L bar K^{-1} mol^{-1} \times 450 K)^{-1} \\ & \Rightarrow K_C=\frac{2.0 \times 10^{10} bar^{-1}}{(0.0831 L bar K^{-1} mol^{-1} \times 450 K)^{-1}} \\ &=(2.0 \times 10^{10} bar^{-1})(0.0831 Lbar K^{-1} mol^{-1} \times 450 K) \\ &=74.79 \times 10^{10} L mol^{-1} \\ &=7.48 \times 10^{11} L mol^{-1} \\ &=7.48 \times 10^{11} M^{-1} \end{aligned} \end{aligned} $

Answer

The initial concentration of $HI$ is $0.2 atm$. At equilibrium, it has a partial pressure of $0.04 atm$. Therefore, a decrease in the pressure of $HI$ is $0.2 \hat{a} E^{\prime \prime} 0.04=0.16$. The given reaction is:

$ 2 HI _{(g)} \longleftrightarrow H _{2(g)}+I _{2(g)} $

Initial conc. $\quad 0.2 atm$

$0 \quad 0$

At equilibrium $\quad 0.04 atm$

$ \begin{aligned} & \frac{0.16}{2} \quad \frac{2.15}{2} \\ & =0.08 atm=0.08 atm \end{aligned} $

Therefore,

$ \begin{aligned} K_p & =\frac{p _{H_2} \times p _{I_2}}{p_HI^{2}} \\ & =\frac{0.08 \times 0.08}{(0.04)^{2}} \\ & =\frac{0.0064}{0.0016} \\ & =4.0 \end{aligned} $

Hence, the value of $K_p$ for the given equilibrium is 4.0 .

Answer

The given reaction is:

$N _{2(g)}+3 H _{2(g)} \longleftrightarrow 2 NH _{3(g)}$

The given concentration of various species is

$[N_2]=\frac{1.57}{20} mol L^{-1} \quad[H_2]=\frac{1.92}{20} mol L^{-1}$

$[NH_3]=\frac{8.13}{20} mol L^{-1}$

Now, reaction quotient $Q_c$ is:

$ \begin{aligned} Q_C & =\frac{[NH_3]^{2}}{[N_2][H_2]^{3}} \\ & =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}} \\ & =2.4 \times 10^{3} \end{aligned} $

Since $Q_C \neq K_C$, the reaction mixture is not at equilibrium.

Again, $Q_C>K_C$. Hence, the reaction will proceed in the reverse direction.

Answer

The balanced chemical equation corresponding to the given expression can be written as:

$4 NO _{(g)}+6 H_2 O _{(g)} \longleftrightarrow 4 NH _{3(g)}+5 O _{2(g)}$

Answer

The given reaction is:

$ \begin{aligned} & H_2 O _{(g)}+CO _{(g)} \longleftrightarrow H _{2(g)}+CO _{2(g)} \\ & \begin{matrix} \text{ Initial conc. } & \frac{1}{10} M & \frac{1}{10} M & 0 & 0 \end{matrix} \\ & \text{ At equilibrium } \quad \frac{1-0.4}{10} M \quad \frac{1-0.4}{10} M \quad \frac{0.4}{10} M \quad \frac{0.4}{10} M \\ & =0.06 M \quad=0.06 M \quad=0.04 M \quad=0.04 M \end{aligned} $

Therefore, the equilibrium constant for the reaction,

$ \begin{aligned} K_C & =\frac{[H_2][CO_2]}{[H_2 O][CO]} \\ & =\frac{0.04 \times 0.04}{0.06 \times 0.06} \\ & =0.444 \text{ (approximately) } \end{aligned} $

Answer

It is given that equilibrium constant $K_C$ for the reaction

$H _{2(g)}+I _{2(g)} \longleftrightarrow 2 HI _{(g)}$ is 54.8.

Therefore, at equilibrium, the equilibrium constant $K_C^{\prime}$ for the reaction

$2 HI _{(g)} \longleftrightarrow H _{2(g)}+I _{2(g)}$ will be $\frac{1}{54.8}$.

$[HI]=0.5 molL^{-1}$

Let the concentrations of hydrogen and iodine at equilibrium be $x molL^{-1}$

$[H_2]=[I_2]=x mol L^{-1}$

Therefore, $\frac{[H_2][I_2]}{[HI]^{2}}=K_C^{\prime}$

$\Rightarrow \frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}$

$\Rightarrow x^{2}=\frac{0.25}{54.8}$

$\Rightarrow x=0.06754$

$x=0.068 molL^{-1}$ (approximately)

Hence, at equilibrium, $[H_2]=[I_2]=0.068 mol L^{-1}$.

Answer

The given reaction is:

Now, we can write, $\frac{[I_2][Cl_2]}{[ICl]^{2}}=K_C$

$\Rightarrow \frac{x \times x}{(0.78-2 x)^{2}}=0.14$

$\Rightarrow \frac{x^{2}}{(0.78-2 x)^{2}}=0.14$

$\Rightarrow \frac{x}{0.78-2 x}=0.374$

$\Rightarrow x=0.292-0.748 x$

$\Rightarrow 1.748 x=0.292$

$\Rightarrow x=0.167$

Hence, at equilibrium,

$[ICl]=[I 2]=$

Answer

Let $p$ be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

Initial conc.

$ C_2 H _{6(g)} \longleftrightarrow C_2 H _{4(g)}+H _{2(g)} $

At equilibrium 4.0atm 0 0

We can write,

$ \begin{aligned} & \frac{p _{C_2 H_4} \times p _{H_2}}{p _{C_2 H_6}}=K_p \\ & \Rightarrow \frac{p \times p}{4.0-p}=0.04 \\ & \Rightarrow p^{2}+0.16-0.04 p \\ & \Rightarrow p^{2}+0.04 p-0.16=0 \\ & \text{ Now }, p=\frac{-0.04 \pm \sqrt{(0.04)^{2}-4 \times 1 \times(-0.16)}}{2 \times 1} \\ & \quad=\frac{-0.04 \pm 0.80}{2} \\ & \quad=\frac{0.76}{2} \quad(\text{ Taking positive value}) \\ & \quad=0.38 \end{aligned} $

Hence, at equilibrium,

$ \begin{aligned} {[C_2 H_6]-4-p } & =4-0.38 \\ & =3.62 atm \end{aligned} $

Answer

(i) Reaction quotient,

$ Q_C=\frac{[CH_3 COOC_2 H_5][H_2 O]}{[CH_3 COOH][C_2 H_5 OH]} $

(ii) Let the volume of the reaction mixture be $V$. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

$ \begin{matrix} \text{ At equilibrium } & \frac{1-0.171}{V} & \frac{0.18-0.171}{V} & \frac{0.171}{V} M & \frac{0.171}{V} M \\ = & \frac{0.829}{V} M & =\frac{0.009}{V} M & \end{matrix} $

Therefore, equilibrium constant for the given reaction is:

$ \begin{aligned} K_C & =\frac{[CH_3 COOC_2 H_5][H_2 O]}{[CH_3 COOH][C_2 H_5 OH]} \\ & =\frac{\frac{0.171}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}}=3.919 \\ = & 3.92(\text{ approximately }) \end{aligned} $

(iii) Let the volume of the reaction mixture be $V$.

Therefore, the reaction quotient is,

$ \begin{aligned} Q_C & =\frac{[CH_3 COOC_2 H_5][H_2 O]}{[CH_3 COOH][C_2 H_5 OH]} \\ & =\frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}} \\ & =0.2037 \\ & =0204 \text{ (approximately) } \end{aligned} $

Since $Q_C<K_C$, equilibrium has not been reached.

Answer

Let the concentrations of both $PCl_3$ and $Cl_2$ at equilibrium be $x$ molL ${ }^{\hat{e} e^{-1}}$. The given reaction is:

concentration of $PCL_5$ was found to be $ 0.5 \times 10^{-1} molL^{-1} $. If value of $K_c$ is $8.3 \times 10 ^{-3}$, what are the

At equilibrium $0.5 \times 10^{-1} mol L^{-1} \quad x mol L^{-1} \quad x mol L^{-1}$

It is given that the value of equilibrium constant, $K_C$ is $8.3 \times 10^{-3}$.

Now we can write the expression for equilibrium as:

$ \begin{aligned} & \frac{[PCl_2][Cl_2]}{[PCl_5]}=K_C \\ & \Rightarrow \frac{x \times x}{0.5 \times 10^{-1}}=8.3 \times 10^{-3} \\ & \Rightarrow x^{2}=4.15 \times 10^{-4} \\ & \Rightarrow x=2.04 \times 10^{-2} \\ & \quad=0.0204 \\ & \quad=0.02 \text{ (approximately) } \end{aligned} $

Therefore, at equilibrium,

$[PCl_3]=[Cl_2]=0.02 mol L{ }^{-1}$.

Answer

For the given reaction,

$ FeO_{(g)} + CO_{(g)} \longleftrightarrow Fe_{(s)} + CO_{2(g)} $

Initialy, $\quad 1.4 atm \quad 0.80 atm$

$ \begin{aligned} Q_p & =\frac{p _{CO_2}}{p _{CO}} \\ & =\frac{0.80}{1.4} \\ & =0.571 \end{aligned} $

It is given that $K_p=0.265$.

Since $Q_P>K_P$, the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of $CO$ will increase while the pressure of $CO_2$ will decrease.

Now, let the increase in pressure of $CO=$ decrease in pressure of $CO_2$ be $p$.

Then, we can write,

$ \begin{aligned} & K_P=\frac{p _{CO_2}}{p _{CO}} \\ & \Rightarrow 0.265=\frac{0.80-p}{1.4+p} \\ & \Rightarrow 0.371+0.265 p=0.80-p \\ & \Rightarrow 1.265 p=0.429 \\ & \Rightarrow p=0.339 atm \end{aligned} $

Therefore, equilibrium partial of $CO_2, p _{CO_2}=0.80-0.339=0.461 atm$.

And, equilibrium partial pressure of $CO, p _{CO}=1.4+0.339=1.739 atm$.

Answer

The given reaction is:

$ N _{2(g)}+3 H _{2(g)} \longleftrightarrow \quad 2 NH _{3(g)} $

At a particular time: $3.0 mol L^{-1} \quad 2.0 mol L^{-1} \quad 0.5 mol L^{-1}$

Now, we know that,

$ \begin{aligned} Q_C & =\frac{[NH_3]^{2}}{[N_2][H_2]^{3}} \\ & =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}} \\ & =0.0104 \end{aligned} $

It is given that $K_C=0.061$.

Since $Q_C \neq K_C$, the reaction is not at equilibrium.

Since $Q_C<K_C$, the reaction will proceed in the forward direction to reach equilibrium.

Answer

Let the amount of bromine and chlorine formed at equilibrium be $x$. The given reaction is:

$2BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)} $

Initail conc. $ 3.3 \times 10^{-3} m$

At equilibrium $ 3.3 \times 10 ^{-2} -2a$

Now, we can write,

$ \begin{aligned} & \frac{[Br_2][Cl_2]}{[BrCl^{2}]}=K_C \\ \Rightarrow & \frac{x \times x}{(3.3 \times 10^{-3}-2 x)^{2}}=32 \\ \Rightarrow & \frac{x}{3.3 \times 10^{-3}-2 x}=5.66 \\ \Rightarrow & x=18.678 \times 10^{-3}-11.32 x \\ \Rightarrow & 12.32 x=18.678 \times 10^{-3} \\ \Rightarrow & x=1.5 \times 10^{-3} \end{aligned} $

Therefore, at equilibrium,

$ \begin{aligned} {[BrCl] } & =3.3 \times 10^{-3}-(2 \times 1.5 \times 10^{-3}) \\ & =3.3 \times 10^{-3}-3.0 \times 10^{-3} \\ & =0.3 \times 10^{-3} \\ & =3.0 \times 10^{-4} molL^{-1} \end{aligned} $

Answer

Let the total mass of the gaseous mixture be $100 g$.

Mass of $CO=90.55 g$

And, mass of $CO_2=(100 a \in " 90.55)=9.45 g$

Now, number of moles of $CO$,

$ n _{CO}=\frac{90.55}{28}=3.234 mol $

Number of moles of $CO_2$

$ n _{CO_2}=\frac{9.45}{44}=0.215 mol $

Partial pressure of $CO$,

$ \begin{aligned} p _{CO} & =\frac{n _{CO}}{n _{CO}+n _{CO_2}} \times p _{\text{total }} \\ & =\frac{3.234}{3.234+0.215} \times 1 \\ & =0.938 atm \end{aligned} $

Partial pressure of $CO_2$,

$ \begin{aligned} p _{CO_2} & =\frac{n _{CO_2}}{n _{CO}+n _{CO_2}} \times p _{\text{total }} \\ & =\frac{0.215}{3.234+0.215} \times 1 \\ & =0.062 atm \end{aligned} $

Therefore, $K_P=\frac{[CO]^{2}}{[CO_2]}$

$ \begin{aligned} & =\frac{(0.938)^{2}}{0.062} \\ & =14.19 \end{aligned} $

For the given reaction,

$\Delta n=2 -1=1$

We know that,

$K_p=K_C(R T)^{\Delta r}$

$\Rightarrow 14.19=K_C(0.082 \times 1127)^{\prime}$

$\Rightarrow K_C=\frac{14.19}{0.082 \times 1127}$

$=0.154$ (approximately)

Answer

(a) For the given reaction,

$\Delta G^{\circ}=\Delta G^{\circ}$ ( Products) â $\epsilon^{\prime \prime} \Delta G^{\circ}$ ( Reactants)

$\Delta G^{\circ}=52.0$ - ${87.0+0}$

= - $35.0 kJ mol|^{\hat{\epsilon} e^{\prime \prime}}$

(b) We know that,

$\Delta G^{\circ}=RT \log K_c$

$\Delta G^{\circ}=2.303 RT \log K_c$

$K_c=\frac{-35.0 \times 10^{-3}}{-2.303 \times 8.314 \times 298}$

$=6.134$

$\therefore K_c=$ antilog (6.134)

$=1.36 \times 10^{6}$

Hence, the equilibrium constant for the given reaction $K_c$ is $1.36 \times 10^{6}$

Answer

(a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

(c) The number of moles of reaction products remains the same.

Answer

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.

The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.

The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.

Answer

Given,

$K _{\text{P for the reaction i.e., }} H _{2(g)}+Br _{2(g)} \longleftrightarrow 2 HBr _{(g)}$ is $1.6 \times 10^{5}$.

Therefore, for the reaction $2 HBr _{(g)} \longleftrightarrow H _{2(g)}+Br _{2(g)}$, the equilibrium constant will be,

$ \begin{aligned} K_P^{\prime} & =\frac{1}{K_P} \\ & =\frac{1}{1.6 \times 10^{5}} \\ & =6.25 \times 10^{-6} \end{aligned} $

Now, let $p$ be the pressure of both $H_2$ and $Br_2$ at equilibrium.

Now, we can write,

$ \begin{aligned} & \frac{p _{HBr} \times p_2}{p_HBr^{2}}=K_P^{\prime} \\ & \frac{p \times p}{(10-2 p)^{2}}=6.25 \times 10^{-6} \end{aligned} $

$\frac{p}{10-2 p}=2.5 \times 10^{-3}$

$p=2.5 \times 10^{-2}-(5.0 \times 10^{-3}) p$

$p+(5.0 \times 10^{-3}) p=2.5 \times 10^{-2}$

$(1005 \times 10^{-3}) p=2.5 \times 10^{-2}$

$p=2.49 \times 10^{-2}$ bar $=2.5 \times 10^{-2}$ bar (approximately)

Therefore, at equilibrium,

$[H_2]=[Br_2]=2.49 \times 10^{-2}$ bar

$[HBr]=10-2 \times(2.49 \times 10^{-2})$ bar

$=9.95 bar=10 bar$ (approximately)

Answer

(a) For the given reaction, $K_P = \frac{p_{CO} \times p^3_{H_2}}{p_{CH_4} \times p_{H_2 O}}$

(b) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Answer

(a) According to Le Chatelier’s principle, on addition of $H_2$, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of $CH_3 OH$, the equilibrium will shift in the backward direction.

(c) On removing $CO$, the equilibrium will shift in the backward direction.

(d) On removing $CH_3 OH$, the equilibrium will shift in the forward direction.

Answer (a)

(b) Value of $K_c$ for the reverse reaction at the same temperature is:

$ \begin{aligned} K_c^{\prime} & =\frac{1}{K_c} \\ & =\frac{1}{8.3 \times 10^{-3}}=1.2048 \times 10^{2} \\ & =120-48 \end{aligned} $

(c) (i) $K_c$ would remain the same because in this case, the temperature remains the same.

(ii) $K_c$ is constant at constant temperature. Thus, in this case, $K_c$ would not change.

(iii) In an endothermic reaction, the value of $K_c$ increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of $K_c$ will increase if the temperature is increased.

Answer

Let the partial pressure of both carbon dioxide and hydrogen gas be $p$. The given reaction is:

$ CO _{(g)}+H_2 O _{(g)} \longleftrightarrow CO _{2(g)}+H _{2(g)} $

$\begin{matrix} \text{ Initial conc. } & 4.0 bar & 4.0 bar & 0\end{matrix} $

$\begin{matrix} \text{ At equilibrium } & 4.0-p & 4.0-p & p\end{matrix} $

It is given that $K_p=10.1$.

Now,

$ \begin{aligned} & \frac{p _{CO_2} \times p _{H_2}}{p _{COH} \times p_2}=K_P \\ & \Rightarrow \frac{p \times p}{(4.0-p)(4.0-p)}=10.1 \\ & \Rightarrow \frac{p}{4.0-p}=3.178 \\ & \Rightarrow p=12.712-3.178 p \\ & \Rightarrow 4.178 p=12.712 \\ & \Rightarrow p=3.04 \end{aligned} $

Hence, at equilibrium, the partial pressure of $H_2$ will be 3.04 bar.

Answer

If the value of $K_c$ lies between $10^{-3}$ and $10^{3}$, a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

Answer

The given reaction is: $3 O _{2(g)} \longleftrightarrow 2 O _{3(g)}$

Then, $K_C=\frac{[O _{3(g)}]^{2}}{[O _{2(g)}]^{3}}$

It is given that $K_C=2.0 \times 10^{-50}$ and $[O _{2(g)}]=1.6 \times 10^{-2}$.

Then, we have,

$2.0 \times 10^{-50}=\frac{[O _{3(g)}]^{2}}{[1.6 \times 10^{-2}]^{3}}$

$\Rightarrow[O _{3(g)}]^{2}=2.0 \times 10^{-50} \times(1.6 \times 10^{-2})^{3}$

$\Rightarrow[O _{3(g)}]^{2}=8.192 \times 10^{-56}$

$\Rightarrow[O _{3(g)}]=2.86 \times 10^{-28} M$

Hence, the concentration of $O_3$ is $2.86 \times 10^{-28} M$.

Answer

Let the concentration of methane at equilibrium be $x$.

$ CO _{(g)}+3 H _{2(g)} \longleftrightarrow CH _{4(g)}+H_2 O _{(g)} $

At equilibrium $\frac{0.3}{1}=0.3 M \quad \frac{0.1}{1}=0.1 M \quad x \quad \frac{0.02}{1}=0.02 M$

It is given that $K_c=3.90$.

Therefore,

$ \begin{gathered} \frac{[CH _{4(g)}][H_2 O _{(g)}]}{[CO _{(g)}][H _{2(g)}]^{3}}=K_c \\ \Rightarrow \frac{x \times 0.02}{0.3 \times(0.1)^{3}}=3.90 \\ \Rightarrow x=\frac{3.90 \times 0.3 \times(0.1)^{3}}{0.02} \\ \quad=\frac{0.00117}{0.02} \\ =0.0585 M \\ =5.85 \times 10^{-2} M \end{gathered} $

Hence, the concentration of $CH_4$ at equilibrium is $5.85 \times 10^{-2} M$.

Answer

A conjugate acid-base pair is a pair that differs only by one proton.

The conjugate acid-base for the given species is mentioned in the table below.

Answer

Lewis acids are those acids which can accept a pair of electrons. For example, $BF_3, H^{+}$, and $NH_4^{+}$are Lewis acids.

Answer

The table below lists the conjugate bases for the given Bronsted acids.

Bronsted acid

$HF$

$H_2 SO_4$

$HCO_3^{-}$

Conjugate base

$Fi$

$HSO_4^{-}$

$CO_3^{2-}$

Answer

The table below lists the conjugate acids for the given Bronsted bases.

Bronsted base

$NH_2^{-}$

$NH_3$

$HCOO^{-}$

Conjugate acid

$NH_3$

$NH_4^{+}$

$HCOOH$

Answer

The table below lists the conjugate acids and conjugate bases for the given species.

Species

$H_2 O$

$HCO_3^{-}$

$HSO_4^{-}$

$NH_3$

Conjugate acid

$H_3 O^{+}$

$H_2 CO_3$

$H_2 SO_4$

$NH_4^{+}$

Conjugate base

$OH^{-}$

$CO_3^{2-}$

$SO_4^{2-}$

$NH_2^{-}$

Answer

(a) $OH^{-}$is a Lewis base since it can donate its lone pair of electrons.

(b) $F^{-}$is a Lewis base since it can donate a pair of electrons.

(c) $H^{+}$is a Lewis acid since it can accept a pair of electrons.

(d) $BCl_3$ is a Lewis acid since it can accept a pair of electrons.

Answer

Given,

$[H^{+}]=3.8 \times 10^{-3} M$

$\therefore pH$ value of soft drink

$=-\log [H^{+}]$

$ \begin{aligned} & =-\log (3.8 \times 10^{-3}) \\ & =-\log 3.8-\log 10^{-3} \\ & =-\log 3.8+3 \\ & =-0.58+3 \\ & =2.42 \end{aligned} $

Answer

Given,

$pH=3.76$

It is known that,

$ \begin{aligned} & pH=-\log [H^{+}] \\ & \Rightarrow \log [H^{+}]=-pH \\ & \Rightarrow[H^{+}]=antilog(-pH) \\ & =\text{ antil og }(-3.76) \\ & =1.74 \times 10^{-4} M \end{aligned} $

Hence, the concentration of hydrogen ion in the given sample of vinegar is $1.74 \times 10^{-4} M$.

Answer

It is known that,

$K_b=\frac{K_w}{K_a}$

Given,

$K_a$ of $HF=6.8 \times 10^{-44}$

Hence, $K_b$ of its conjugate base $F^{-}$ $=\frac{K_w}{K_a}$

$=\frac{10^{-14}}{6.8 \times 10^{-4}}$

$=1.5 \times 10^{-11}$

Given,

$K_a$ of $HCOOH=1.8 \times 10^{-4 }$

Hence, $K_b$ of its conjugate base $HCOO^{-}$

$=\frac{K_w}{K_a}$

$=\frac{10^{-14}}{1.8 \times 10^{-4}}$

$=5.6 \times 10^{-11}$

Given,

$K _{\text{a }}$ of $HCN=4.8 \times 10^{-9}$

Hence, $K_b$ of its conjugate base $CN^{\text{ate- }}$

$=\frac{K_w}{K_a}$

$=\frac{10^{-14}}{4.8} \times 10^{-9}$

$=2.08 \times 10^{-6}$

Answer

Ionization of phenol:

As the value of the ionization constant is very less, $x$ will be very small. Thus, we can ignore $x$ in the denominator.

$ \begin{aligned} \therefore x & =\sqrt{1 \times 10^{-10} \times 0.05} \\ & =\sqrt{5 \times 10^{-12}} \\ & =2.2 \times 10^{-6} M=[H_3 O^{+}] \end{aligned} $

Since $[H_3 O^{+}]=[C_6 H_5 O^{-}]$,

$[C_6 H_5 O^{-}]=2.2 \times 10^{-6} M$.

Now, let $\alpha$ be the degree of ionization of phenol in the presence of $0.01 M C_6 H_5 ONa$.

$ C_6 H_5 ONa \longrightarrow C_6 H_5 O^{-}+Na^{+} $

Conc.

$ 0.01 $

Also,

$ \begin{aligned} & \quad C_6 H_5 OH+H_2 O \longleftrightarrow C_6 H_5 O^{-}+H_3 O^{+} \\ & \text{Conc. } 0.05-0.05 \alpha \\ & {[C_6 H_5 OH]=0.05-0.05 \alpha ; 0.05 M} \end{aligned} $

$ \begin{aligned} & {[C_6 H_5 O^{-}]=0.01+0.05 \alpha ; 0.01 M} \\ & {[H_3 O^{+}]=0.05 \alpha} \\ & K_a=\frac{[C_6 H_5 O^{-}][H_3 O^{+}]}{[C_6 H_5 OH]} \\ & K_a=\frac{(0.01)(0.05 \alpha)}{0.05} \\ & 1.0 \times 10^{-10}=.01 \alpha \\ & \alpha=1 \times 10^{-8} \end{aligned} $

Answer

(i) To calculate the concentration of $HS^{-}$ ion:

Case I (in the absence of $HCl$ ):

Let the concentration of $HS^{-}$ be $x M$.

$ \begin{aligned} & H_2 S \longleftrightarrow H^{+}+HS^{-} \\ & \begin{matrix} C_i & 0.1 & 0 & 0 \end{matrix} \\ & \begin{matrix} C_f & 0.1-x & x & x \end{matrix} \\ & \text{ Then, } K _{a_1}=\frac{[H^{+}][HS^{-}]}{[H_2 S]} \\ & 9.1 \times 10^{-8}=\frac{(x)(x)}{0.1-x} \\ & (9.1 \times 10^{-8})(0.1-x)=x^{2} \end{aligned} $

Taking $0.1-x M ; 0.1 M$, we have $(9.1 \times 10^{-8})(0.1)=x^{2}$.

$ \begin{aligned} & 9.1 \times 10^{-9}=x^{2} \\ & x=\sqrt{9.1 \times 10^{-9}} \\ & \quad=9.54 \times 10^{-5} M \\ & \Rightarrow[HS^{-}]=9.54 \times 10^{-5} M \end{aligned} $

Case II (in the presence of $HCl$ ):

In the presence of $0.1 M$ of $HCl$, let $[HS^{-}]$be $y M$.

Then, $\quad H_2 S \longleftrightarrow HS^{-}+H^{+}$

$\begin{matrix} C_i & 0.1 & 0 & 0\end{matrix} $

$C_f \quad 0.1-y \quad y \quad y$

Also, $\quad HCl \longleftrightarrow H^{+}+Cl^{-}$

$0.1 \quad 0.1$

Now, $K _{a_1}=\frac{[HS^{-}][H^{+}]}{[H_2 S]}$

$K_{a_1}=\frac{[y] (0.1+y)}{(0.1-y)}$

$9.1 \times 10^{-8}=\frac{y \times 0.1}{0.1} \quad(\because 0.1-y ; 0.1 M)$ (and $0.1+y ; 0.1 M$ )

$9.1 \times 10^{-8}=y$

$\Rightarrow[HS^{-}]=9.1 \times 10^{-8}$

(ii) To calculate the concentration of $[S^{2-}]$ :

Case I (in the absence of $0.1 M HCl$ ):

$HS^{-} \longleftrightarrow H^{+}+S^{2-}$

$[HS^{-}]=9.54 \times 10^{-5} M$ (From first ionization, case I)

Let $[S^{2-}]$ be $X$.

Also, $[H^{+}]=9.54 \times 10^{-5} M$ (From first ionization, case I)

$K _{a_2}=\frac{[H^{+}][S^{2-}]}{[HS^{-}]}$

$K _{a_2}=\frac{(9.54 \times 10^{-5})(X)}{9.54 \times 10^{-5}}$

$1.2 \times 10^{-13}=X=[S^{2-}]$

Case II (in the presence of $0.1 M HCl$ ):

Again, let the concentration of $HS^{-}$ be $X^{\prime} M$.

$ \begin{aligned} & {[HS^{-}]=9.1 \times 10^{-8} M \text{ (From first ionization, case II) }} \\ & {[H^{+}]=0.1 M _{(\text{From } HCl, \text{ case II) }}} \end{aligned} $

$ \begin{aligned} & {[S^{2-}]=X^{\prime}} \\ & \text{ Then, } K _{a_2}=\frac{[H^{+}][S^{2-}]}{[HS^{-}]} \\ & 1.2 \times 10^{-13}=\frac{(0.1)(X^{\prime})}{9.1 \times 10^{-8}} \\ & 10.92 \times 10^{-21}=0.1 X^{\prime} \\ & \frac{10.92 \times 10^{-21}}{0.1}=X^{\prime} \\ & X^{\prime}=\frac{1.092 \times 10^{-20}}{0.1} \\ & =1.092 \times 10^{-19} M \\ & \Rightarrow K _{a_1}=1.74 \times 10^{-5} \end{aligned} $

Answer

Method 1

1. $\quad CH_3 COOH \longleftrightarrow CH_3 COO^{-}+H^{+} \quad K_a=1.74 \times 10^{-5}$
2. $H_2 O+H_2 O \longleftrightarrow H_3 O^{+}+OH^{-} \quad K_w=1.0 \times 10^{-14}$

Since $K a \gg K_w$ :

$ \begin{matrix} CH_3 COOH+H_2 O \longleftrightarrow CH_3 COO^{-}+H_3 O^{+} \\ 0.05 & 0 & 0 \\ 0.05-.05 \alpha & 0.05 \alpha & 0.05 \alpha \\ K_a & =\frac{(.05 \alpha)(.05 \alpha)}{(.05-0.05 \alpha)} \\ & =\frac{(.05 \alpha)(0.05 \alpha)}{.05(1-\alpha)} \\ & =\frac{.05 \alpha^{2}}{1-\alpha} \end{matrix} $

$ \begin{aligned} & 1.74 \times 10^{-5}=\frac{0.05 \alpha^{2}}{1-\alpha} \\ & 1.74 \times 10^{-5}-1.74 \times 10^{-5} \alpha=0.05 \alpha^{2} \\ & 0.05 \alpha^{2}+1.74 \times 10^{-5} \alpha-1.74 \times 10^{-5} \\ & D=b^{2}-4 a c \\ & =(1.74 \times 10^{-5})^{2}-4(.05)(1.74 \times 10^{-5}) \\ & =3.02 \times 10^{-25}+.348 \times 10^{-5} \\ & \alpha=\sqrt{\frac{K_a}{c}} \\ & \alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}} \\ & =\sqrt{\frac{34.8 \times 10^{-5} \times 10}{10}} \\ & =\sqrt{3.48 \times 10^{-6}} \\ & =CH_3 COOH \longleftrightarrow CH_3 COO^{-}+H^{+} \end{aligned} $

$\alpha \underline{1.86 \times 10^{-3}}$

$ \begin{aligned} {[CH_3 COO^{-}] } & =0.05 \times 1.86 \times 10^{-3} \\ & =\frac{0.93 \times 10^{-3}}{1000} \\ & =.000093 \end{aligned} $

Method 2

Degree of dissociation,

$\alpha=\sqrt{\frac{K_a}{c}}$

$c=0.05 M$

$K_a=1.74 \times 10^{-5}$

Then, $\alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}}$

$ \begin{aligned} & \alpha=\sqrt{34.8 \times 10^{-5}} \\ & \alpha=\sqrt{3.48} \times 10^{-4} \\ & \alpha=1.8610^{-2} \end{aligned} $

$CH_3 COOH \longleftrightarrow CH_3 COO^{-}+H^{+}$

Thus, concentration of $CH_3 COO^-=$ c.a

$ \begin{aligned} & =.05 \times 1.86 \times 10^{-2} \\ & =.093 \times 10^{-2} \\ & =.00093 M \\ & \text{ Since }[oAc^{-}]=[H^{+}] \\ & {[H^{+}]=.00093=.093 \times 10^{-2} .} \\ & pH=-\log [H^{+}] \\ & =-\log (.093 \times 10^{-2}) \\ & \therefore pH=3.03 \end{aligned} $

Hence, the concentration of acetate ion in the solution is $0.00093 M$ and its $Ph$ is 3.03 .

Answer

Let the organic acid be HA.

$\Rightarrow HA \longleftrightarrow H^{+}+A^{-}$

Concentration of $HA=0.01 M$

$pH=4.15$

$-\log [H^{+}]=4.15$

$[H^{+}]=7.08 \times 10^{-5}$

$K_a=\frac{[H^{+}][A^{-}]}{[HA]}$

$[H^{+}]=[A^{-}]=7.08 \times 10^{-5}$

$[HA]=0.01$

Then,

$ \begin{aligned} K_a & =\frac{(7.08 \times 10^{-5})(7.08 \times 10^{-5})}{0.01} \\ K_a & =5.01 \times 10^{-7} \\ p K_a & =-\log K_a \\ & =-\log (5.01 \times 10^{-7}) \\ p K_a & =6.3001 \end{aligned} $

Answer

(i) $0.003 MHCl$ :

$H_2 O+HCl \longleftrightarrow H_3 O^{+}+Cl^{-}$

Since $HCl$ is completely ionized,

$ \begin{aligned} & {[H_3 O^{+}]=[HCl] .} \\ & \Rightarrow[H_3 O^{+}]=0.003 \end{aligned} $

Now,

$ \begin{aligned} pH & =-\log [H_3 O^{+}]=-\log (.003) \\ & =2.52 \end{aligned} $

Hence, the $pH$ of the solution is 2.52 .

(ii) $0.005 MNaOH$ :

$ \begin{aligned} & NaOH _{(a q)} \longleftrightarrow Na _{(a q)}^{+}+HO _{(a q)}^{-} \\ & {[HO^{-}]=[NaOH]} \\ & \Rightarrow[HO^{-}]=.005 \\ & pOH=-\log [HO^{-}]=-\log (.005) \\ & pOH=2.30 \\ & \therefore pH=14-2.30 \\ & \quad=11.70 \end{aligned} $

Hence, the $pH$ of the solution is 11.70 .

(iii) $0.002 HBr$ :

$ \begin{aligned} & HBr+H_2 O \longleftrightarrow H_3 O^{+}+Br^{-} \\ & {[H_3 O^{+}]=[HBr]} \\ & \Rightarrow[H_3 O^{+}]=.002 \\ & \therefore pH=-\log [H_3 O^{+}] \\ & \quad=-\log (0.002) \\ & \quad=2.69 \end{aligned} $

Hence, the $pH$ of the solution is 2.69 .

(iv) $0.002 M KOH$ :

$ \begin{aligned} & KOH _{(a q)} \longleftrightarrow K _{(a q)}^{+}+OH _{(a q)}^{-} \\ & {[OH^{-}]=[KOH]} \\ & \Rightarrow[OH^{-}]=.002 \\ & \text{ Now, pOH }=-\log [OH^{-}] \\ & \quad=2.69 \\ & \begin{matrix} \therefore pH=14-2.69 \\ =11.31 \end{matrix} \end{aligned} $

Hence, the $pH$ of the solution is 11.31 .

Answer

(a) For $2 g$ of $TIOH$ dissolved in water to give $2 L$ of solution:

$ \begin{aligned} {[TlOH _{(a q)}] } & =\frac{2}{2} g / L \\ & =\frac{2}{2} \times \frac{1}{221} M \\ & =\frac{1}{221} M \end{aligned} $

$TlOH _{(a q)} \longrightarrow Tl _{(a q)}^{+}+OH _{(a q)}^{-}$

$[OH _{(a q)}^{-}]=[TlOH _{(o q)}]=\frac{1}{221} M$

$K_w=[H^{+}][OH^{-}]$

$10^{-14}=[H^{+}] (\frac{1}{221})$

$221 \times 10^{-14}=[H^{+}]$

$ \begin{aligned} \Rightarrow pH=-\log [H^{+}] & =-\log (221 \times 10^{-14}) \\ & =-\log (2.21 \times 10^{-12}) \\ & =11.65 \end{aligned} $

(b) For $0.3 g$ of $Ca(OH)_2$ dissolved in water to give $500 mL$ of solution:

$ \begin{aligned} & Ca(OH)_2 \longrightarrow Ca^{2+}+2 OH^{-} \\ & {[Ca(OH)_2]=0.3 \times \frac{1000}{500}=0.6 M} \\ & {[OH^{-} _{a q}]=2 \times[Ca(OH) _{2 a q}]=2 \times 0.6} \\ & =1.2 M \\ & {[H^{+}]=\frac{K_w}{[OH _{a q}^{-}]}} \\ & =\frac{10-14}{1.2} M \\ & =0.833 \times 10^{-14} \\ & pH=-\log (0.833 \times 10^{-14}) \\ & =-\log (8.33 \times 10^{-13}) \\ & =(-0.902+13) \\ & =12.098 \end{aligned} $

(c) For $0.3 g$ of $NaOH$ dissolved in water to give $200 mL$ of solution: $NaOH \longrightarrow Na _{(a q)}^{+}+OH _{(a q)}^{-}$

$[NaOH]=0.3 \times \frac{1000}{200}=1.5 M$

$[OH^{-} _{a q}]=1.5 M$

Then, $[H^{+}]=\frac{10^{-14}}{1.5}$

$ =6.66 \times 10^{-13} $

$pH=-\log (6.66 \times 10^{-13})$

$ =12.18 $

(d) For $1 mL$ of $13.6 M HCl$ diluted with water to give $1 L$ of solution:

$13.6 \times 1 mL=M_2 \times 1000 mL$

(Before dilution) (After dilution)

$13.6 \times 10^{-43}=M_2 \times 1 L$

$M_2=1.36 \times 10^{-2}$

$[H^{+}]=1.36 \times 10^{\text{-2 }}$

$pH=- \log (1.36 \times 10^{-2})$

$=(- 0.1335+2)$

$=1.866 - 41.87$

Answer

Degree of ionization, $\alpha=0.132$

Concentration, $c=0.1 M$

Thus, the concentration of $H_3 O^{+}=c . \alpha$

$=0.1 \times 0.132$

$=0.0132$

$ \begin{aligned} pH & =-\log [H^{+}] \\ & =-\log (0.0132) \\ & =1.879: 1.88 \end{aligned} $

Now,

$ \begin{aligned} K_a & =C \alpha^{2} \\ & =0.1 \times(0.132)^{2} \\ K_a & =.0017 \\ p K_a & =2.75 \end{aligned} $

Answer

$c=0.005$

$pH=9.95$

$pOH=4.05$

$pH=- \log (4.105)$

$4.05=-\log [OH^{-}]$

$[OH^{-}]=8.91 \times 10^{-5}$

$c \alpha=8.91 \times 10^{-5}$

$\alpha=\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}=1.782 \times 10^{-2}$

Thus, $K_b=c \alpha^{2}$

$ \begin{aligned} & =0.005 \times(1.782)^{2} \times 10^{-4} \\ & =0.005 \times 3.1755 \times 10^{-4} \\ & =0.0158 \times 10^{-4} \end{aligned} $

$K_b=1.58 \times 10^{-6}$

$P k_b=-\log K_b$

$=-\log (1.58 \times 10^{-6})$

$=5.80$

Answer

$K_b=4.27 \times 10^{-10}$

$c=0.001 M$

$pH=$ ?

$\alpha=$ ?

$k_b=c \alpha^{2}$

$4.27 \times 10^{-10}=0.001 \times \alpha^{2}$

$4270 \times 10^{-10}=\alpha^{2}$

$65.34 \times 10^{-5}=\alpha=6.53 \times 10^{-4}$

Then, $[$ anion $]=c \alpha=.001 \times 65.34 \times 10^{-5}$

$ =.065 \times 10^{-5} $

$pOH=-\log (.065 \times 10^{-5})$

$ =6.187 $

$pH=7.813$

Now,

$K_a \times K_b=K_w$

$\therefore 4.27 \times 10^{-10} \times K_a=K_w$

$K_a=\frac{10^{-14}}{4.27 \times 10^{-10}}$

$ =2.34 \times 10^{-5} $

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{- 5}$.

Answer

$ \begin{aligned} & c=0.05 M \\ & p K_a=4.74 \\ & p K_a=-\log (K_a) \\ & K_a=1.82 \times 10^{-5} \\ & K_a=c \alpha^{2} \quad \alpha=\sqrt{\frac{K_a}{c}} \\ & \alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}=1.908 \times 10^{-2} \end{aligned} $

When $HCl$ is added to the solution, the concentration of $H^{+}$ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When $0.01 M HCl$ is taken.

Let $x$ be the amount of acetic acid dissociated after the addition of $HCl$.

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 - $x$ and $0.01+x$ can be taken as 0.05 and 0.01 respectively.

$ \begin{aligned} K_a & =\frac{[CH_3 COO^{-}][H^{+}]}{[CH_3 COOH]} \\ \therefore K_a & =\frac{(0.01) x}{0.05} \\ x & =\frac{1.82 \times 10^{-5} \times 0.05}{0.01} \\ x & =1.82 \times 10^{-3} \times 0.05 M \end{aligned} $

Now,

$ \begin{aligned} & =\frac{1.82 \times 10^{-3} \times 0.05}{0.05} \\ & =1.82 \times 10^{-3} \end{aligned} $

$ \alpha=\frac{\text{ Amount of acid dissociated }}{\text{ Amount of acid taken }} $

Case II: When $0.1 M HCl$ is taken.

Let the amount of acetic acid dissociated in this case be $X$. As we have done in the first case, the concentrations of various species involved in the reaction are:

$ \begin{aligned} & {[CH_3 COOH]=0.05-X ; 0.05 M} \\ & {[CH_3 COO^{-}]=X} \\ & {[H^{+}]=0.1+X ; 0.1 M} \\ & K_a=\frac{[CH_3 COO^{-}][H^{+}]}{[CH_3 COOH^{-}]} \\ & \therefore K_a=\frac{(0.1) X}{0.05} \\ & x=\frac{1.82 \times 10^{-5} \times 0.05}{0.1} \\ & x=1.82 \times 10^{-4} \times 0.05 M \end{aligned} $

Now,

$ \begin{aligned} \alpha & =\frac{\text{ Amount of acid dissociated }}{\text{ Amount of acid taken }} \\ & =\frac{1.82 \times 10^{-4} \times 0.05}{0.05} \\ & =1.82 \times 10^{-4} \end{aligned} $

Answer

$K_b=5.4 \times 10^{-4}$

$c=0.02 M$

Then, $\alpha=\sqrt{\frac{K_b}{c}}$

$ \begin{aligned} & =\sqrt{\frac{5.4 \times 10^{-4}}{0.02}} \\ & =0.1643 \end{aligned} $

Now, if $0.1 M$ of $NaOH$ is added to the solution, then $NaOH$ (being a strong base) undergoes complete ionization.

$ \begin{aligned} & NaOH _{(a q)} \longleftrightarrow Na _{(a q)}^{+}+OH _{(a q)}^{-} \\ & 0.1 M \quad 0.1 M \end{aligned} $

And, $(CH_3)_2 NH$ $H_2 O$ $\longleftrightarrow$ $(0.02-x)$ $; 0.02 M$ $(CH_3)_2 NH_2^{+}+\overline{O} H$ $x$ $x$ $; 0.1 M$

Then, $[(CH_3)_2 NH_2^{+}]=x$

$[OH^{-}]=x+0.1 ; 0.1$

$\Rightarrow K_b=\frac{[(CH_3)_2 NH_2^{+}][OH^{-}]}{[(CH_3)_2 NH]}$

$5.4 \times 10^{-4}=\frac{x \times 0.1}{0.02}$

$x=0.0054$

It means that in the presence of $0.1 M NaOH, 0.54 %$ of dimethylamine will get dissociated.

Answer

(a) Human muscle fluid 6.83:

$pH=6.83$

$pH=-\log [H^{+}]$

$\therefore 6.83=-\log [H^{+}]$

$[H^{+}]=1.48 \times 10^{-7} M$

(b) Human stomach fluid, 1.2:

$pH=1.2$

$1.2=-\log [H^{+}]$

$\therefore[H^{+}]=0.063$

(c) Human blood, 7.38:

$pH=7.38=-\log [H^{+}]$ $\therefore[H^{+}]=4.17 \times 10^{-8} M$

(d) Human saliva, 6.4:

$pH=6.4$

$6.4=-\log [H^{+}]$

$[H^{+}]=3.98 \times 10^{-7}$

Answer

The hydrogen ion concentration in the given substances can be calculated by using the given relation:

$pH=-\log [H^{+}]$

(i) $pH$ of milk $=6.8$

Since, $pH=a - \log [H^{+}]$

$6.8=- \log [H^{+}]$

$\log [H^{+}]=$- 6.8

$[H^{+}]=anitlog(-8)$

$=1.5 \times 10^{-7} M$

(ii) $pH$ of black coffee $=5.0$

Since, $pH=a - \log [H^{+}]$

$5.0=\hat{a} - \log [H^{+}]$

$\log [H^{+}]=$- 5.0

$[H^{+}]=anitlog(-5.0)$

$=10^{-5} M$

(iii) $pH$ of tomato juice $=4.2$

Since, $pH=- \log [H^{+}]$

$4.2= - \log [H^{+}]$

$\log [H^{+}]=-{4.2}$

$[H^{+}]=anitlog(4.2)$

$=6.31 \times 10^{-5} M$

(iv) $pH$ of lemon juice $=2.2$

Since, $pH=- \log [H^{+}]$ $2.2=- \log [H^{+}]$

$\log [H^{+}]=- 2.2$

$[H^{+}]=anitlog(- 2)$

$=6.31 \times 10^{-3} M$

(v) $pH$ of egg white $=7.8$

Since, $pH=- \log [H^{+}]$

$7.8=- \log [H^{+}]$

$\log [H^{+}]=- 7.8$

$[H^{+}]=anitlog(-7.8)$

$=1.58 \times 10^{-8} M$

Answer

$[KOH _{a q}]=\frac{0.561}{\frac{1}{5}} g / L$

$=2.805 g / L$

$=2.805 \times \frac{1}{56.11} M$

$=.05 M$

$KOH _{(a q)} \to K _{(a q)}^{+}+OH^{-} _{(a q)}$

$[OH^{-}]=.05 M=[K^{+}]$

$[H^{+}][H^{-}]=K_w$

$[H^{+}] \frac{K_w}{[OH^{-}]}$

$=\frac{10^{-14}}{0.05}=2 \times 10^{-13} M$

$\therefore pH=12.70$

Answer

Solubility of $Sr(OH)_2=19.23 g / L$

Then, concentration of $Sr(OH)_2$

$=\frac{19.23}{121.63} M$

$=0.1581 M$

$Sr(OH) _{2(a q)} \longrightarrow Sr^{2+} _{(a q)}+2(OH^{-}) _{(a q)}$

$\therefore[Sr^{2+}]=0.1581 M$

$[OH^{-}]=2 \times 0.1581 M=0.3126 M$

Now,

$K_w=[OH^{-}][H^{+}]$

$\frac{10^{-14}}{0.3126}=[H^{+}]$

$\Rightarrow[H^{+}]=3.2 \times 10^{-14}$

$\therefore pH=13.495 \approx 13.50$

Answer

Let the degree of ionization of propanoic acid be $\alpha$.

Then, representing propionic acid as HA, we have:

$ \begin{aligned} & HA+H_2 O \longleftrightarrow H_3 O^{+}+A^{-} \\ & (.05-0.0 \alpha) \approx .05 \end{aligned} $

$ \begin{gathered} .05 \alpha \\ K_a=\frac{[H_3 O^{+}][A^{-}]}{[HA]} \\ \quad=\frac{(.05 \alpha)(.05 \alpha)}{0.05}=.05 \alpha^{2} \\ \alpha=\sqrt{\frac{K_a}{.05}}=1.63 \times 10^{-2} \\ \text{ Then, }[H_3 O^{+}]=.05 \alpha=.05 \times 1.63 \times 10^{-2}=K_b .15 \times 10^{-4} M \\ \therefore pH=3.09 \end{gathered} $

In the presence of $0.1 M$ of $HCl$, let $\alpha^{\prime}$ be the degree of ionization.

$ \begin{aligned} & \text{ Then, }[H_3 O^{+}]=0.01 \\ & {[A^{-}]=005 \alpha^{\prime}} \\ & {[HA]=.05} \\ & K_a=\frac{0.01 \times .05 \alpha^{\prime}}{.05} \\ & 1.32 \times 10^{-5}=.01 \times \alpha^{\prime} \\ & \alpha^{\prime}=1.32 \times 10^{-3} \end{aligned} $

Answer

$c=0.1 M$

$pH=2.34$ $-\log [H^{+}]=pH$

$-\log [H^{+}]=2.34$

$[H^{+}]=4.5 \times 10^{-3}$

Also,

$[H^{+}]=c \alpha$

$4.5 \times 10^{-3}=0.1 \times \alpha$

$\frac{4.5 \times 10^{-3}}{0.1}=\alpha$

$\alpha=45 \times 10^{-3}=.045$

Then,

$K_a=c \alpha^{2}$

$=0.1 \times(45 \times 10^{-3})^{2}$

$=202.5 \times 10^{-6}$

$=2.02 \times 10^{-4}$

Answer

$NaNO_2$ is the salt of a strong base $(NaOH)$ and a weak acid $(HNO_2)$.

$NO_2^{-}+H_2 O \longleftrightarrow HNO_2+OH^{-}$

$K_h=\frac{[HNO_2][OH^{-}]}{[NO_2^{-}]}$

$\Rightarrow \frac{K_w}{K_a}=\frac{10^{-14}}{4.5 \times 10^{-4}}=.22 \times 10^{-10}$

Now, If $x$ moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: $[NO_2^{-}]=.04-x ; 0.04$

$[HNO_2]=x$

$[OH^{-}]=x$

$K_h=\frac{x^{2}}{0.04}=0.22 \times 10^{-10}$

$x^{2}=.0088 \times 10^{-10}$

$x=.093 \times 10^{-5}$

$\therefore[OH^{-}]=0.093 \times 10^{-5} M$

$[H_3 O^{+}]=\frac{10^{-14}}{.093 \times 10^{-5}}=10.75 \times 10^{-9} M$

$\Rightarrow pH=-\log (10.75 \times 10^{-9})$

$=7.96$

Therefore, degree of hydrolysis

$=\frac{x}{0.04}=\frac{.093 \times 10^{-5}}{.04}=2.325 \times 10^{- 5}$

Answer

$pH=3.44$

We know that,

$pH=- \log [H^{+}]$

$\therefore[H^{+}]=3.63 \times 10^{-4}$

Then, $K_h=\frac{(3.63 \times 10^{-4})^{2}}{0.02} \quad(\because$ concentration $=0.02 M)$

$\Rightarrow K_h=6.6 \times 10^{-6}$

Now, $K_h=\frac{K_w}{K_a}$

$\Rightarrow K_a=\frac{K_w}{K_h}=\frac{10^{-14}}{6.6 \times 10^{-6}}$

$=1.51 \times 10^{-9}$

Answer

(i) $NaCl$ :

$ NaCl+H_2 O \longleftrightarrow \underset{\text{ Strong base }}{NaOH}+\underset{\text{ Strong acid }}{HCl} $

Therefore, it is a neutral solution.

(ii) $KBr$ :

$ KBr + H _2O \longleftrightarrow \underset{\text{ Strong base }}{KOH} + \underset{\text{ Strong acid }}{HBr} $

Therefore, it is a neutral solution.

(iii) $NaCN$ :

$ \begin{aligned} & NaCN+H_2 O \longleftrightarrow HCN+NaOH \\ & \text{ Weak acid Strong base } \end{aligned} $

Therefore, it is a basic solution.

(iv) $NH_4 NO_3$

$ NH_4 NO_3+H_2 O \longleftrightarrow \underset{\text{ Weak base }}{NH_4 OH+\underset{\text{ Strong acid }}{HNO_3}} $

Therefore, it is an acidic solution.

(v) $NaNO_2$

$ NaNO_2+H_2 O \longleftrightarrow \underset{\text{ Strong base }}{NaOH}+\underset{\text{ Weak acid }}{HNO_2} $

Therefore, it is a basic solution.

(vi) $KF$

$ \begin{aligned} & KF+H_2 O \longleftrightarrow KOH+HF \\ & \text{ Strong base Weak acid } \end{aligned} $

Therefore, it is a basic solution.

Answer

It is given that $K_a$ for $ClCH_2 COOH$ is $1.35 \times 10^{-33}$.

$ \begin{aligned} & \Rightarrow K_a=c \alpha^{2} \\ & \therefore \alpha=\sqrt{\frac{K_a}{c}} \\ & \quad=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}} \quad(\therefore \text{ concentration of acid }=0.1 m) \\ & \begin{aligned} \alpha & =\sqrt{1.35 \times 10^{-2}} \\ & =0.116 \\ \therefore[H^{+}] & =c \alpha=0.1 \times 0.116 \\ \quad & =.0116 \\ \Rightarrow pH & =-\log [H^{+}]=1.94 \end{aligned} \end{aligned} $

$ClCH_2 COONa$ is the salt of a weak acid i.e., $ClCH_2 COOH$ and a strong base i.e., $NaOH$.

$ \begin{aligned} & ClCH_2 COO^{-}+H_2 O \longleftrightarrow ClCH_2 COOH+OH^{-} \\ & K_h=\frac{[ClCH_2 COOH][OH^{-}]}{[ClCH_2 COO^{-}]} \\ & K_h=\frac{K_w}{K_a} \\ & K_h=\frac{10^{-14}}{1.35 \times 10^{-3}} \\ & =0.740 \times 10^{-11} \\ & \text{ Also, } .K_h=\frac{x^{2}}{0.1} \quad (\text{ where } x \text{ is the concentration of } OH^{-} \text{and } ClCH_2 COOH) \\ & 0.740 \times 10^{-11}=\frac{x^{2}}{0.1} \\ & 0.074 \times 10^{-11}=x^{2} \\ & \Rightarrow x^{2}=0.74 \times 10^{-12} \\ & \begin{matrix} x=0.86 \times 10^{-6} \\ {[OH^{-}]=0.86 \times 10^{-6}} \\ \therefore[H^{+}]=\frac{K_w}{0.86 \times 10^{-6}} \\ \quad=\frac{10^{-14}}{0.86 \times 10^{-6}} \\ {[H^{+}]=1.162 \times 10^{-8}} \\ pH=-\log [H^{+}] \\ =7.94 \end{matrix} \end{aligned} $

Answer

Ionic product, $K_w=[H^{+}][OH^{-}]$

Let $[H^{+}]=x$.

Since $[H^{+}]=[OH^{-}], K_w=x^{2}$.

$\Rightarrow K_w$ at $310 K$ is $2.7 \times 10^{-14}$.

$\therefore 2.7 \times 10^{-14}=x^{2}$

$\Rightarrow x=1.64 \times 10^{-7}$

$\Rightarrow[H^{+}]=1.64 \times 10^{-7}$

$\Rightarrow pH=-\log [H^{+}]$

$=-\log [1.64 \times 10^{-7}]$

$=6.78$

Hence, the $pH$ of neutral water is 6.78 .

Answer

(a)

Moles of $H_3 O^{+}=\frac{25 \times 0.1}{1000}=.0025 mol$

Moles of $OH^{-}=\frac{10 \times 0.2 \times 2}{1000}=.0040 mol$

Thus, excess of $OH^{-}=.0015 mol$

$ \begin{aligned} & {[OH^{-}]=\frac{.0015}{35 \times 10^{-3}} mol / L=.0428} \\ & \begin{aligned} pOH & =-\log [OH] \\ & =1.36 \\ pH & =14-1.36 \\ & =12.63 \quad \text{ (not matched) } \end{aligned} \end{aligned} $

Moles of $H_3 O^{+}=\frac{2 \times 10 \times 0.01}{1000}=.0002 mol$

Moles of $OH^{-}=\frac{2 \times 10 \times .01}{1000}=.0002 mol$

Since there is neither an excess of $H_3 O^{+}$or $OH^{-}$, the solution is neutral. Hence, $pH=7$.

(c)

$ \text{Moles of } H _3O^+ = \frac{2\times 10\times 0.1}{1000} =.002 mol $

Moles of $OH^{-}=\frac{10 \times 0.1}{1000}=0.001 mol$

Excess of $H_3 O^{+}=.001 mol$

Thus, $[H_3 O^{+}]=\frac{.001}{20 \times 10^{-3}}=\frac{10^{-3}}{20 \times 10^{-3}}=.05$

$\therefore pH=-\log (0.05)$

$=1.30$

Answer

(1) Silver chromate:

$ Ag_2 CrO_4 \longrightarrow 2 Ag^{+}+CrO_4{ }^{2-} $

Then,

$K _{s p}=[Ag^{+}]^{2}[CrO_4{ }^{2-}]$

Let the solubility of $Ag_2 CrO_4$ be s.

$\Rightarrow[Ag^{+}] 2 s$ and $[CrO_4{ }^{2-}]=s$

Then,

$K _{s p}=(2 s)^{2} . s=4 s^{3}$

$\Rightarrow 1.1 \times 10^{-12}=4 s^{3}$

$.275 \times 10^{-12}=s^{3}$

$s=0.65 \times 10^{-4} M$

Molarity of $Ag^{+}=2 s=2 \times 0.65 \times 10^{{-4}}=1.30 \times 10^{{-4}} M$

Molarity of $CrO_4{ }^{2-}=s=0.65 \times 10^{- 4} M$

(2) Barium chromate:

$BaCrO_4 \longrightarrow Ba^{2+}+CrO_4{ }^{2-}$

Then, $K _{s p}=[Ba^{2+}][CrO_4{ }^{2-}]$

Let $s$ be the solubility of $BaCrO_4$.

Thus, $[Ba^{2+}]=s$ and $[CrO_4^{2-}]=s$

$\Rightarrow K _{S P}=s^{2}$

$\Rightarrow 1.2 \times 10^{-10}=s^{2}$

$\Rightarrow s=1.09 \times 10^{-5} M$

Molarity of $Ba^{2+}=$ Molarity of $CrO_4{ }^{2-}=s=1.09 \times 10^{-5} M$

(3) Ferric hydroxide:

$Fe(OH)_3 \longrightarrow Fe^{2+}+3 OH^{-}$

$K _{s p}=[Fe^{2+}][OH^{-}]^{3}$

Let $s$ be the solubility of $Fe(OH)_3$.

Thus, $[Fe^{3+}]=s$ and $[OH^{-}]=3 s$

$\Rightarrow K _{s p}=s .(3 s)^{3}$

$ =s .27 s^{3} $

$K _{s p}=27 s^{4}$

$1.0 \times 10^{-38}=27 s^{4}$

$.037 \times 10^{-38}=s^{4}$

$.00037 \times 10^{-36}=s^{4} \quad \Rightarrow 1.39 \times 10^{-10} M=S$

Molarity of $Fe^{3+}=s=1.39 \times 10^{-10} M$

Molarity of $OH^{-}=3 s=4.17 \times 10^{-10} M$

(4) Lead chloride:

$PbCl_2 \longrightarrow Pb^{2+}+2 Cl^{-}$

$K _{S P}=[Pb^{2+}][Cl^{-}]^{2}$

Let $K _{S P}$ be the solubility of $PbCl_2$. $[PB^{2+}]=s$ and $[Cl^{-}]=2 s$

Thus, $K _{s p}=s .(2 s)^{2}$

$=4 s^{3}$

$\Rightarrow 1.6 \times 10^{-5}=4 s^{3}$

$\Rightarrow 0.4 \times 10^{-5}=s^{3}$

$4 \times 10^{-6}=s^{3} \Rightarrow 1.58 \times 10^{-2} M=S .1$

Molarity of $PB^{2+}=s=1.58 \times 10^{-2} M$

Molarity of chloride $=2 s=3.16 \times 10^{-2} M$

(5) Mercurous iodid

Answer

Let $s$ be the solubility of $Ag_2 CrO_4$.

Then, $Ag_2 CrO_4 \longleftrightarrow Ag^{2+}+2 CrO_4^{-}$

$K _{s p}=(2 s)^{2} . s=4 s^{3}$

$1.1 \times 10^{-12}=4 s^{3}$

$s=6.5 \times 10^{-5} M$

Let $s^{\prime}$ be the solubility of AgBr.

$AgBr _{(s)} \longleftrightarrow Ag^{+}+Br^{-}$

$K _{s p}=s^{\prime 2}=5.0 \times 10^{-13}$

$\therefore s^{\prime}=7.07 \times 10^{-7} M$

Therefore, the ratio of the molarities of their saturated solution is $\frac{s}{s^{\prime}}=\frac{6.5 \times 10^{-5} M}{7.07 \times 10^{-7} M}=91.9$.

Answer

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., $0.001 M$.

Then,

$ \begin{aligned} & NalO_3 \to Na^{+}+10_3^{-} \\ & 0.001 M \quad 0.001 M \\ & Cu(ClO_3)_2 \to Cu^{2+}+2 ClO_3^{-} \\ & 0.001 M \quad 0.001 M \end{aligned} $

Now, the solubility equilibrium for copper iodate can be written as:

$ Cu(10_3)_2 \to Cu^{2+} _{(aq)}+210 _{3(aq)}^{-} $

Ionic product of copper iodate:

$ \begin{aligned} & =[Cu^{2+}][10_3^{-}]^{2} \\ & =(0.001)(0.001)^{2} \\ & =1 \times 10^{-9} \end{aligned} $

Since the ionic product $(1 \times 10^{{-9}})$ is less than $K _{s p}(7.4 \times 10^{- 8})$, precipitation will not occur.

Answer

Since $pH=3.19$,

$ [H_3 O^{+}]=6.46 \times 10^{-4} M $

$C_6 H_5 COOH+H_2 O \longleftrightarrow C_6 H_5 COO^{-}+H_3 O$

$K_a \frac{[C_6 H_5 COO^{-}][H_3 O^{+}]}{[C_6 H_5 COOH]}$

$ \frac{[C_6 H_5 COOH]}{[C_6 H_5 COO^{-}]}=\frac{[H_3 O^{+}]}{K_a}=\frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}}=10 $

Let the solubility of $C_6 H_5 COOAg$ be $x mol / L$.

Then,

$ \begin{aligned} & {[Ag^{+}]=x} \\ & {[C_6 H_5 COOH]+[C_6 H_5 COO^{-}]=x} \\ & 10[C_6 H_5 COO^{-}]+[C_6 H_5 COO^{-}]=x \\ & {[C_6 H_5 COO^{-}]=\frac{x}{11}} \\ & K _{s p}[Ag^{+}][C_6 H_5 COO^{-}] \\ & 2.5 \times 10^{-13}=x(\frac{x}{11}) \\ & x=1.66 \times 10^{-6} mol / L \end{aligned} $

Thus, the solubility of silver benzoate in a pH 3.19 solution is $ 1.66 \times 10^{-6} molL^{-1} $

Now, let the solubility of $C_6 H_5 COOAg$ be $x^{\prime} mol / L$.

Then, $[Ag^{+}]=x^{\prime} M$ and $[CH_3 COO^{-}]=x^{\prime} M$.

$K _{s p}=[Ag^{+}][CH_3 COO^{-}]$

$K _{s p}=(x^{\prime})^{2}$

$x^{\prime}=\sqrt{K _{s p}}=\sqrt{2.5 \times 10^{-13}}=5 \times 10^{-7} mol / L$

$\therefore \frac{x}{x^{\prime}}=\frac{1.66 \times 10^{-6}}{5 \times 10^{-7}}=3.32$

Hence, $C_6 H_5 COOAg$ is approximately 3.317 times more soluble in a low $pH$ solution.

Answer

Let the maximum concentration of each solution be $x mol / L$. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., $\frac{x}{2}$.

$ \therefore[FeSO_4]=[Na_2 S]=\frac{x}{2} M $

Then, $[Fe^{2+}]=[FeSO_4]=\frac{x}{2} M$

Also, $[S^{2-}]=[Na_2 S]=\frac{x}{2} M$

$FeS _{(s)} \longleftrightarrow Fe^{2+} _{(a q)}+S^{2-} _{(a q)}$

$K _{s p}=[Fe^{2+}][S^{2-}]$

$6.3 \times 10^{-18}=(\frac{x}{2})(\frac{x}{2})$

$\frac{x^{2}}{4}=6.3 \times 10^{-18}$

$\Rightarrow x=5.02 \times 10^{-9}$

If the concentrations of both solutions are equal to or less than $5.02 \times 10^{ae 9} M$, then there will be no precipitation of iron sulphide.

Answer

$CaSO _{4(s)} \longleftrightarrow Ca^{2+} _{(aq)}+ SO _{4(aq)}^{2-} $

$K _{s p}=[Ca^{2+}][SO_4^{2-}]$

Let the solubility of $CaSO_4$ be $s$.

Then, $K _{s p}=s^{2}$

$9.1 \times 10^{-6}=s^{2}$

$s=3.02 \times 10^{-3} mol / L$

Molecular mass of $CaSO_4=136 g / mol$

Solubility of $CaSO_4$ in gram $/ L=3.02 \times 10^{-3} \times 136$

$=0.41 g / L$

This means that we need $1 L$ of water to dissolve $0.41 g$ of $CaSO_4$

Therefore, to dissolve $1 g$ of $CaSO_4$ we require $=\frac{1}{0.41} L=2.44 L$ of water.

Answer

For precipitation to take place, it is required that the calculated ionic product exceeds the $K _{s p}$ value.

Before mixing:

$ [S^{2-}]=1.0 \times 10^{-19} M[M^{2+}]=0.04 M $

volume $=10 mL \quad$ volume $=5 mL$

After mixing:

$ [S^{2-}]=? \quad[M^{2+}]=? $

volume $=(10+5)=15 mL \quad$ volume $=15 mL$

$ \begin{aligned} & {[S^{2-}]=\frac{1.0 \times 10^{-19} \times 10}{15}=6.67 \times 10^{-20} M} \\ & {[M^{2+}]=\frac{0.04 \times 5}{15}=1.33 \times 10^{-2} M} \end{aligned} $

$ \begin{aligned} \text{ Ionic product } & =[M^{2+}][S^{2-}] \\ & =(1.33 \times 10^{-2})(6.67 \times 10^{-20}) \\ & =8.87 \times 10^{-22} \end{aligned} $

This ionic product exceeds the $K _{s p}$ of $Zns$ and $CdS$. Therefore, precipitation will occur in $CdCl_2$ and $ZnCl_2$ solutions.