50 ml of 0.11 $CH_3 COOH$ is titraqted with 0.1 M NaOH

0 ml of $NaOH is added to the weak acid solution

• $[H^+]=\sqrt{K_a \times C_{a c i d}}$

10 ml of $NaOH is added to the salt$CH_3COONa+CH_3COOH$

$CH_3COONa+CH_3COOH$ is buffer solution

$pH=p K_a+\log \frac{[{ salt }]}{{ acid }}$

At the equivalence point

50 ml of 0.1 M NaOH +50 ml of $0.1 \mathrm{M}$ acetic acid

$Concentration =\frac{no.of mole of salt formed}{total volume}$

$Concentration=\frac{5 {mmol of salt }}{(100 {ml})} =.05 {mol}{L}$

m mol= Vol in m l $\times$ molarity

Molarity = m mol/vol in ml

Titration of weak base with strong acid

$NH_4OH$ with HCl

No acid is added

$OH^{-}$ $=$ $\sqrt{K_b\times {[base]}}$

When we add HCl before equivalence point

$pOH=PK_b+\log\frac{[salt]}{[{ base }]}$

At the equivalence point salt of weak base with strong acid

$[H^+] =\sqrt{K_h \times C_{{salt }}}$

$[H^+]=\sqrt{\frac{K_w}{K_b} \times { [salt }]}$

After the equivalent point

salt + strong acid

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

Salts are of three types

Salts which are non-electrolyte

Soluble salt $\xrightarrow {H_2O}$ breaks (complete dissociation)

Example $NaCl \longrightarrow Na^{+} (aq + Cl^{-} (aq)$

Sparingly soluble salt - lot of salts are sparingly soluble

$\quad$ Undissociated $\quad \quad \quad \quad \quad$ dissociated

Undissociated and dissociated species are in equilibrium

So, the concept of ionic equilibrium can be applied

$A g C l(s) \rightleftharpoons A g^{+}(a q)+Cl^{-}(a q)$

Solubility of salt is S mol/l

$AgCl(s) \rightleftharpoons Ag^{+}(a q)+Cl^{-}(a q)$

$\quad \quad \quad \quad \quad$ S mol/L of $Ag^{+} \quad$ S mol/L of $Cl^-$

$AgCl(s) \rightleftharpoons A g^{+}(aq)+Cl^{-}(aq)$

$K=\frac{\left[\mathrm{Ag}^{\dagger}\right]\left[\mathrm{Cl}^{-}\right]}{\operatorname{AgCl}(\mathrm{s}) \sim 1}$

$K_{sp}=[{Ag}^{+}][{Cl}^{-}]$

$K_{sp}$ is solubility product

$AgCl (s) \rightleftharpoons Ag^+ (aq) Cl^- (aq)$

S mol/L of AgCl goes to solution that means solubility is S mol/l

• $[Ag^+]= S mol / L$

$[Cl^-]=S mol / L$

$K_{s p}=[Ag^+][Cl^-]$

$K_{s p}$= S$\times$ S =$S^2$

$S=\sqrt {K_{SP}}$

This will be solubility of the salt

Problem The solubility of $A_2 X_3$ is y mol $dm^{-3}$.Its solubility product is

Solution

• $A_2X_3 \rightleftharpoons2 A^{3+}+3 X^{2-}$

• $\quad\quad\quad\quad$2y $\quad\quad\quad$ 3y

• $K_{s p}=[A^{3+}]^2[X^{2-}]^3$

• $K_{s p}=(2 y)^2(3 y)^3=4 y^2 \times 27 y^3$

• $K_{s p}=108 y^5$

$AB(s) \rightleftharpoons A^+ (a q) + B^- (a q)$

$K_{s p}= S \times S=S^2$

$A B_2 \rightleftharpoons A^{2+}+2 B^-$

$\quad\quad\quad\quad S\quad 2S$

• $K_{s p} =(S)^1(2S)^2$

• $K_{s p} =S \times 4S^2$

• $K_{s p}=4S^3$

Problem For a sparingly soluble salt ApBq, the relationship of solubility product (Ls) and its solubility is

Solution

$A_p B_q \rightleftharpoons p A^{q +}+qB ^{P-}$

$\quad\quad\quad\quad\quad\ pS\quad\quad qS$

$Ls=[A^{q+}]^p[B^{p-}]^q$

$Ls=(pS)^p(qS)^q$

$Ls=p^p q^q S^{p+q}$

Problem The molar solubility (in mol $L^{-1}$ )of a sparingly soluble salt $MX_4$ is ' s '. The corresponding solubility product is Ksp. 's' is given in terms of Ksp by the relation

Solution

• $MX_4 \rightleftharpoons M^{+4}(a q)+4 X^{-}$

$\quad \quad \quad \quad \quad \quad S \quad \quad 4S$

• $K_{s p}=[M^{4+}] [X^{-}]^4$

• $K_{s p}=S \times(4S)^4 = {4 \times 4 \times 4 \times 4 S^5}$

• $K_{s p}=256S^5$

Problem The solubility of calcium sulphate in water is 4.9 $\times$ $10^{-3} mol {dm}^{-3}$ at 298 K. Calculate the value of Ksp for $CaSO_4$ at this temperature

Solution

• $CaSO_4 \rightleftharpoons Ca^{2+}+SO_4^{2-}$

$\quad \quad \quad \quad \quad \quad S \quad \quad S$

• $S = 4.9 \times 10^{-3}$

• $K_{s p} =S \times S=S^2=\left(4.9 \times 10^{-3}\right)^2$

Problem The solubility product of a salt having general formula $MX_2$, in water is : $4 \times 10^{-12}$.The concentration of $M^{2+}$ ions in the aqueous solution of the salt is

Solution

• $MX_2 \rightleftharpoons M^{2 +}+2 X^{-}$

$\quad \quad \quad \quad \quad S \quad \quad S$

• $K_{s p}=S \times(2S)^2=S \times 4 S^2=4S^3$

• $4 \times 10^{-12}=4 \times S^3$

• $S=1 \times 10^{-4}$

Common ion effect

MX (s) $\rightleftharpoons M^+ (aq)+X^- (aq)$

AgCl ( sparingly soluble salt)

• $AgCl(s) \rightleftharpoons Ag + (aq)+Cl^-(aq)$

$AgNO_3 \rightarrow Ag^{+}+NO_3^{-}$

$\quad\quad\quad\quad\quad\quad\quad\quad2S \quad\quad \quad S$

• $K_{SP}=[Ag+]^2[CrO_4^{2-}] =(2S)^2 \times S$

• $K_{SP} =4S^3$

• $4S^3 = 1.1 \times 10^{-12}$

Solution

• $Ag_2 CrO_4 \rightleftharpoons 2 Ag^{+}+CrO_4^{2-}$

• $AgNO_3 \longrightarrow Ag^++NO_3^{-}$

$\quad\quad\quad\quad\quad\quad 0.1 M \quad 0.1 M$

• $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$

s= y mol/L $\quad\quad$ 2y+0.1 $\quad\quad$ y

$\quad\quad\quad\quad\quad\quad \approx$ 0.1

Solution

• $K_{SP_{Ag_2CrO_4}}$=$[Ag^{+}]^2[CrO_4^{2-}]$

• $1.1 \times 10^{-12}=(0.1)^2 \times y$

• $y=\frac{1.1\times 10^{-12}}{[0.1]^2}$

• $y=1.1 \times 10^{-10} mol / L$

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

• $S^3 =.25 \times 10^{-12}$

• $S \approx 15 \times 10^{-4}$

Ionic product and solubility product for sparingly soluble salt like AgCl

$AgCl \rightleftharpoons [Ag^{+}][Cl^{-}]$

• $[Ag^{+}][Cl^{-}]$ = $K_{salt}$ (Ionic product)

AgCl $\rightleftharpoons Ag^+ + Cl^-$

$\quad\quad\quad\quad$ In solution

Multiplication of $[Ag^{+}][Cl^{-}]$ in a saturated solution is a known as solubility product

Concept of solubility product can be used to know under what condition a precipitate will form

Ionic product < solubility product then solution is not saturated

Ionic product = solubility product then solution is saturated

Further addition of salt will lead to precipitation

Ionic product > solubility product then the solution is oversaturated

Problem Solid Ba(NO₃)₂ is gradually dissolved in a $1\times 10^{-4}M Na_2NO_3$ solution.At what concentration of $Ba^{2+}$ will a precipitate begin to form

$(K_{sp}$ for $BaCO_3=5.1\times 10^{-9})$

Solution

$Na_2 CO_3 \longrightarrow 2 Na^{+}+CO_3^{2-}$

• $BaCO_3$ (sparingly soluble)

Solution

$BaCO_3 \rightleftharpoons Ba^{2+}+CO_3^{2-}$

At what concentration of $Ba^{2+}$, the ppt. will start to form

• $BaCO_3 \Longrightarrow K_{s p}=[Ba^{2+}][{CO}_3^2]$

Solution

• $[CO_3^{2-}] \simeq[Na_2 CO_3]$

• $[CO_3^{2-}]=1 \times 10^{-4}$ M

• $K_{SP}=5.1 \times 10^{-9}=[Ba^{2+}][CO_3^{2-}]$

• $[Ba^{2+}] =\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}$

• $[Ba^{2+}]=5.1 \times 10^{-5}M$

Problem At $25^{\circ} {C}$, the solubility product of $Mg(OH)_2$ is $1.1\times 10^{-11}$. At which pH, will

$Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001 M Mg^{2+}$

ions ?

Solution

• $K_{s p}=K_w$

• $K_{SP}=[Mg^{2+}][OH^{-}]^2$

• $1.1 \times 10^{-11}=[Mg^{2+}][OH^{-}]^2$

• $[OH^-]^2=\frac{1.1 \times 10^{-11}}{.001}$

$[OH^-]^2 = 1.1 \times 10^{-8}$

• $[OH^-] \approx 1 \times 10^{-4}$

Solution

pOH=4

pH=10

At pH- 10 ,$Mg^{2+}$ will start to precipitate

• $K_{SP}=K_w$

• $K_w>K_{SP}$

Problem Calculate the molar solubility of $Mg(OH)_2$ in $1 MNH_4Cl$

• $K_{sp}Mg(OH)_2=1.8 \times 10^{-11}$

• $K_b(NH_3)=1.8 \times 10^{-5}$

Solution

• $NH_4 Cl \rightarrow NH_4^{+}+Cl^{-}$

$\quad 1M\quad\quad \quad\quad M$

• $NH_4^{+}+OH^{-} \rightleftharpoons NH_3+H_2O$

Solution

• $Mg(OH)_2 \rightleftharpoons Mg^{2+}+2 OH^{-}$

• $NH_{4}^++OH^{-} \rightleftharpoons NH_3+H_2O$

• $NH_3+H_2 O \rightleftharpoons NH_{4}^+ + OH^{-}$

• $K_b=1.8 \times 10^{-5}$

Solution

Suppose S is the solubility of $Mg(OH)_2$ in 1M $NH_4Cl$

• $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-}$

$\quad\quad\quad\quad\quad\quad\quad S \quad\quad\quad\quad 2S-X$

• $NH_4^{+} + OH^- \rightleftharpoons NH_3 + H_2O$

$1 M \quad\quad2S \quad\quad \quad0\quad\quad\quad \quad 0$

$1-x \quad 2S-x \quad\quad x \quad\quad\quad x$

Solution

• $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-}$

$\quad \quad \quad \quad\quad \quad \quad S\quad\quad \quad 2S-x$

• $NH_4^{+} + OH^- \rightleftharpoons NH_3 + H_2O$

$\quad1-x \quad 2S-x \quad \quad \quad x$

• $\quad \approx1$

• $K_{SP} = S\times [2S-x]^2$

• $K_b=\frac{[NH_4^+[OH^-]}{[NH_3]}$

• $=\frac{1 \times[OH^-]}{x}$

The concentration of $[OH^-]$ and S (solubility of salt) can be solved now

Precipitation is the basis of wet chemistry

If we want to know the presence of particular ion then u should check there is any salt of that ion which is insoluble

If we know that a salt of that particular ion is sparingly soluble then we will add the counter ion of the particular ion which make this ion into insoluble salt

insoluble salt will get precipitated out

The particular ion presence can be known

If $Ag^+$ is added to a solution and we see a precipitate

It is infer that $Cl^- \quad Br^- \quad I^-$ is present