chemistry-class-11-unit-07-chapter-06-ionic-equillibrium-l-6-8-nriy79suvrs

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Equilibrium concept </primary>
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- Dissociation of weak acid
- $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}($aq $)+\mathrm{H}^{+}$(aq)
- $K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}$
- $K_a$ = acid dissociation constant

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<footer style = "font-size: 18px">Dissociation of molecules $\rightarrow$ Dissociation of molecules $\rightarrow$ <span style = "color:blue"> Equilibrium concept </span> $\rightarrow$ Equilibrium concept $\rightarrow$ Equilibrium concept </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Equilibrium concept </primary>
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- Dissociation of weak bases
- $\mathrm{NH}_3 \stackrel{+\mathrm{H}_2 \mathrm{O}}{\rightleftharpoons}$
- $\mathrm{NH}_4^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$
- $K_b=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}_4^{-}\right]}{\left[\mathrm{NH}_3\right]}$
- $K_b$ = base dissociation constant

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<footer style = "font-size: 18px">Dissociation of molecules $\rightarrow$ Equilibrium concept $\rightarrow$ <span style = "color:blue"> Equilibrium concept </span> $\rightarrow$ Equilibrium concept $\rightarrow$ Degree of dissociation </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Degree of dissociation </primary>
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* Moles of acid,base or salt which exist in ionic form per mole of acid, base or salt

* 1 mole of $CH_3COOH$
- $CH_3COOH \xtofrom {H_2O} CH_3COOH^- (aq) + H^+ (aq)$
- $1 \quad \quad \quad \quad \quad  \quad\quad \quad 0\quad\quad \quad \quad\quad\quad \quad\quad \quad\quad0 $
- $(1-\alpha)\quad \quad \quad \quad  \quad \alpha \quad \quad\quad\quad\quad\quad \quad\quad\quad\quad\alpha$

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<footer style = "font-size: 18px">Equilibrium concept $\rightarrow$ Equilibrium concept $\rightarrow$ <span style = "color:blue"> Degree of dissociation </span> $\rightarrow$ Degree of dissociation $\rightarrow$ Degree of dissociation </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Degree of dissociation </primary>
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- $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
- $C \quad\quad \quad \quad\quad\quad 0 \quad \quad \quad\quad\quad \quad 0$
- $C-C\alpha$$\quad \quad\quad$$C\alpha$ $\quad \quad \quad\quad\quad $$C\alpha$
- $C(1-\alpha)$$ \quad\quad$$C\alpha$ $\quad\quad\quad\quad\quad\quad    $$C\alpha$
- $K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}$
- $K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha)}$

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<footer style = "font-size: 18px">Equilibrium concept $\rightarrow$ Degree of dissociation $\rightarrow$ <span style = "color:blue"> Degree of dissociation </span> $\rightarrow$ Degree of dissociation $\rightarrow$ Degree of dissociation </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Degree of dissociation </primary>
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- $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
- $C(1-\alpha)\quad \quad C\alpha \quad\quad\quad C\alpha$
- $K_a=\frac{C \alpha \times C \alpha}{C(1-\alpha)}$
- $K_a=\frac{C^2 \alpha^2}{C(1-\alpha)}$
- $1 > > \alpha ,(1-\alpha) \simeq 1$
- $K_a=\frac{c^2 \alpha^2}{C \times 1}= C\alpha^2$

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<footer style = "font-size: 18px">Degree of dissociation $\rightarrow$ Degree of dissociation $\rightarrow$ <span style = "color:blue"> Degree of dissociation </span> $\rightarrow$ Degree of dissociation $\rightarrow$ Degree of dissociation </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Degree of dissociation </primary>
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- $NH_3+H_2O \rightleftharpoons NH_4^+(aq)+OH^-(aq)$
- $C(1-\alpha)\quad\quad\quad C\alpha \quad\quad\quad\quad C\alpha$ 
- $K_b=\frac{C_\alpha \times C_\alpha}{C(1-\alpha)}$
- $K_b=\frac{C^2 \alpha^2}{C}=C\alpha^2$
- $\alpha=\sqrt{\frac{K_b}{c}}$
- $[OH^-]=C\alpha=\sqrt{K_b \times C}$

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<footer style = "font-size: 18px">Degree of dissociation $\rightarrow$ Degree of dissociation $\rightarrow$ <span style = "color:blue"> Degree of dissociation </span> $\rightarrow$ Salt hydrolysis $\rightarrow$ Salt hydrolysis </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Salt hydrolysis </primary>
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- **Dissociation of water**
- $H_2 O \rightleftharpoons H^+ + OH^-$
- $H_2 O + H_2 O \rightleftharpoons H_3 O^+ + OH^-$
- $K=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\left[\mathrm{OH}^{-}\right]\right.}{\left[\mathrm{H}_2 \mathrm{O}\right]^2}$
- $K\left[\mathrm{H}_2 O\right]^2=K_W=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]$

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<footer style = "font-size: 18px">Degree of dissociation $\rightarrow$ Degree of dissociation $\rightarrow$ <span style = "color:blue"> Salt hydrolysis </span> $\rightarrow$ Salt hydrolysis $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **Question** $pH$ of $10^{-8} {M}$ HCl

- **Solution**
- $HCl \rightarrow H^+ + Cl^-$
- $10^{-8} \quad 10^-8 M$
- $H_2 O \rightleftharpoons H^+ + OH^-$
- $\quad\quad\quad\quad10^{-7}$
- $H^+ \approx 10^{-7}+10^{-8}$
- $H^+\approx 10^{-7}(1+.1)$
- $p H=-\log H^{+}$

- The $H^+$ ion from water can only be neglected when the concentration of acid or base is quite greater thamn $10^{-7}$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Polyprotic acid $\rightarrow$ Salt </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Salt of weak acid and strong base </primary>
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- $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH+OH^-$
- $C(1-\alpha)\quad \quad \quad \quad \quad  \quad C\alpha\quad \quad \quad \quad  \quad C\alpha$
- $K_W=\frac{[CH_3COOH][OH^-]}{[CH_3COO^{-}]}$
- $K_W=\frac {C \alpha \times C\alpha} {C(1-\alpha)}$

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<footer style = "font-size: 18px">Soluble salt $\rightarrow$ Salt of weak acid and strong base $\rightarrow$ <span style = "color:blue"> Salt of weak acid and strong base </span> $\rightarrow$ Salt of weak acid and strong base $\rightarrow$ Salt of weak acid and strong base </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Salt of weak acid and strong base </primary>
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- $K_h=\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}$
- $K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
- $K_h \times K_a=\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}$ $\times$ $\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
- $K_h \times K_a=K_w$

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<footer style = "font-size: 18px">Salt of weak acid and strong base $\rightarrow$ Salt of weak acid and strong base $\rightarrow$ <span style = "color:blue"> Salt of weak acid and strong base </span> $\rightarrow$ Salt of weak acid and strong base $\rightarrow$ Salt of weak acid and strong base </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-6 </Success>
### <primary style="font-size:24px"> Salt of strong acid and weak base </primary>
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- $NH_3^{+}+H_2O \rightleftharpoons NH_3(aq)+H_3 O^+ $
- $C_{salt} \quad\quad\quad\quad\quad\quad 0\quad \quad\quad\quad 0$
- $C_{salt}(1-\alpha) \quad\quad\quad C\alpha \quad \quad\quad C\alpha$
- $K_h =\frac {K_W}{K_b}=\frac {C^2 \alpha^2}{C(1-\alpha)}=C \alpha^2$
- $C^2 \alpha^2=K_h \times C$
- $[H^{+}]^2=K_h \times C,[H^{+}]=\sqrt{K_h \times C}$

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<footer style = "font-size: 18px">Salt of strong acid and weak base $\rightarrow$ Salt of strong acid and weak base $\rightarrow$ <span style = "color:blue"> Salt of strong acid and weak base </span> $\rightarrow$ Salt of strong acid and weak base $\rightarrow$ Salt of weak acid and strong base </footer>