OSTWALD DILUTION LAW :

• Dissociation constant of weak acid $\left(\mathrm{K}_{\mathrm{a}}\right)$,

$\quad \quad \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{[\mathrm{C} \alpha][\mathrm{C} \alpha]}{\mathrm{C}(1-\alpha)}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}$

$\quad \quad $ If $\alpha < < 1, \text{ then } 1 - \alpha \cong 1 \text{ or }$

$\quad \quad K_a = c \alpha^2 \text{ or } \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{K_a \times V} $

• Similarly for a weak base, $\alpha=\sqrt{\frac{K_{b}}{C}}$. Higher the value of $K_{a} / K_{b}$, strong is the acid / base.

Acidity and $\mathrm{pH}$ scale :

$\therefore \quad pH=-\log a_{H^+} $ (where $ a_{H^+} $ is the activity of $H^+$ ions = molar concentration for dilute solution).

$\mathrm{pH}$ can also be negative or $>14$

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] ; \quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}$

$\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right] ; \quad\left[\mathrm{OH}^{-}\right]=10^{-\mathrm{pOH}}$

$\mathrm{pK_a}=-\log \mathrm{K_a} ; \quad \mathrm{K_a}=10^{-\mathrm{pK_a}}$

$\mathrm{pK_b}=-\log \mathrm{K_b} ; \quad \mathrm{K_b}=10^{-\mathrm{pK_b}}$

PROPERTIES OF WATER :

1. In pure water $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right] \quad$ so it is Neutral.

2. Molar concentration $/$ Molarity of water $=55.56 \mathrm{M}$

3. Ionic product of water $\left(\mathrm{K}_{\mathrm{w}}\right)$ :

$\quad \quad\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$ at $25^{\circ}$ (experimentally)

$\quad \quad\mathrm{pH}=7=\mathrm{pOH} \quad \Rightarrow \quad$ neutral

$\quad \quad\mathrm{pH}<7$ or $\mathrm{pOH}>7 \quad \Rightarrow \quad$ acidic

$\quad \quad\mathrm{pH}>7$ or $\mathrm{pOH}<7 \quad \Rightarrow \quad$ Basic

4. Degree of dissociation of water :

$\quad \quad\alpha=\frac{\text { no. of moles dissociated }}{\text { Total No. of molesinitiallytaken }}=\frac{10^{-7}}{55.55}=$

$\quad \quad 18 \times 10^{-10}$ or $1.8 \times 10^{-7} $%

5. Absolute dissociation constant of water :

$ \quad \quad K_a = K_b =\frac{[H^+][OH^-]}{[H_{2} O]}=\frac{10^{-7} \times 10^{-7}}{55.55}=1.8 \times 10^{-16}$

$\quad \quad pK_a = pK_b =-\log(1.8 \times 10^{-16})=16-\log 1.8=15.74 $

$\quad \quad K_a \times K_{b} =\left[H^{+}][OH^{-}\right]=K_{w}$

$\quad \quad \Rightarrow$ for a conjugate acid- base pairs

$\quad \quad pK_a+pK_b=pK_w=14 \quad at \quad 25^{\circ}C.$.

$\quad \quad p K_{a}$ of ${H}_{3}{O}^{+}$ ions $ =-1.74$

$\quad \quad p K_{b}$ of $\mathrm{OH}^{-}$ ions $=-1.74$.

pH Calculations of Different Types of Solutions:

(a) Strong acid solution :

$\quad$ (i) If concentration is greater than $10^{-6} \mathrm{M}$

$\quad$ In this case $\mathrm{H}^{+}$ ions coming from water can be neglected,

$\quad$(ii) If concentration is less than $10^{-6} \mathrm{M}$

$\quad$ In this case $\mathrm{H}^{+}$ ions coming from water cannot be neglected

(b) Strong base solution :

Using similar method as in part (a) calculate first $\left[\mathrm{OH}^{-}\right]$ and then use

$\left[\mathrm{H}^{+}\right] \times\left[\mathrm{OH}^{-}\right]=10^{-14}$

(c) pH of mixture of two strong acids :

Number of $\mathrm{H}^{+}$ ions from I-solution $=N_1 V_1$

Number of $\mathrm{H}^{+}$ ions from II-solution $=N_2 V_2$

$$ [H^{+}]=N=\frac{N_1V_1 + N_2 V_2}{V_1+V_2} $$

(d) pH of mixture of two strong bases :

$$ [OH^{-}]=N=\frac{N_1V_1 + N_2 V_2}{V_1+V_2} $$

(e) $\mathrm{pH}$ of mixture of a strong acid and a strong base :

If $N_{1} V_{1}>N_{2} V_{2}$, then solution will be acidic in nature and

$$ [H^{+}]=N=\frac{N_1V_1 - N_2 V_2}{V_1+V_2} $$

If $N_{2} V_{2}>N_{1} V_{1}$, then solution will be basic in nature and

$$ [OH^{-}]=N=\frac{N_2 V_2 - N_1 V_1}{V_1+V_2} $$

(f) $\mathrm{pH}$ of a weak acid(monoprotic) solution :

$\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{HA}]}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}$

$if \quad \alpha «1 \Rightarrow(1-\alpha) \approx 1 \quad \Rightarrow \quad K_a \approx C \alpha^{2}$

$\Rightarrow \alpha = \sqrt{\frac{K_a}{C}} ( \text {is valid if } \alpha < 0.1 \text{ or } 10 % )$

On increasing the dilution

$\Rightarrow \mathrm{C} \downarrow \Rightarrow \alpha \uparrow \quad$ and $\left[\mathrm{H}^{+}\right] \downarrow \mathrm{pH} \uparrow$

RELATIVE STRENGTH OF TWO ACIDS :

$\frac{[H^{+}] \quad\text {furnished by I acid }} {[H^{+}] \quad\text {furnished by II acid }} = \frac{C_1 \alpha_{1}}{C_2 \alpha_{2}}=\sqrt{\frac{k_{a_1} c_1}{k_{a_2} c_2}}$

SALT HYDROLYSIS:

Hydrolysis of polyvalent anions or cations

For $[Na_3 PO_4]=C$.

$K_{a1} \times K_{h3}=K_w$

$K_{a1} \times K_{h2}=K_w$

$K_{a3} \times K_{h1}=K_w$

Generally $\mathrm{pH}$ is calculated only using the first step Hydrolysis

$\mathrm{K}_{\mathrm{h} 1}=\frac{\mathrm{Ch}^{2}}{1-\mathrm{h}} \approx \mathrm{Ch}^{2}$

$h=\sqrt{\frac{K_{h_1}}{c}}$

$\Rightarrow\left[\mathrm{OH}^{-}\right]=\mathrm{ch}=\sqrt{\mathrm{K}_{\mathrm{h}_1} \times \mathrm{c}} $

$\Rightarrow[H^{+}]=\sqrt{\frac{K_W \times K_{a_3}}{C}}$

So $pH=\frac{1}{2}[pK_w + pK_{a_3} + \log C]$

BUFFER SOLUTION

(a) Acidic Buffer : e.g. $CH_3 COOH $ and $CH_3 COONa$. (weak acid and salt of its conjugate base).

$$ pH=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { Salt }]}{[\text { Acid }]} \quad \text { [Henderson’s equation ]} $$

(b) Basic Buffer : e.g. $NH_4 OH + NH_4 Cl$. (weak base and salt of its conjugate acid).

$$ \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} $$

SOLUBILITY PRODUCT

$\mathrm{K}_{\mathrm{SP}}=(\mathrm{xs})^{\mathrm{x}}(\mathrm{ys})^{\mathrm{y}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot(\mathrm{s})^{\mathrm{x}+\mathrm{y}}$

CONDITION FOR PRECIPITATION

If ionic product $ K_{I.P} > K_{SP}$ precipitation occurs,

if $K_{I.P}=K_{SP}$ saturated solution (precipitation just begins or is just prevented).

Indicators in AcID Base Titration

For most acid base titrations, it is possible to select indicators which exhibit colour change at $\mathrm{pH}$ close to the equivalence point.

Phenolpthalein

Phenolpthalein is a weak acid, therefore it does not dissociate in the acidic medium and remains in the unionised form, which is colourless.

$$\mathrm{HPh} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Ph}^{-}$$

• In the acidic medium, equilibrium lies to the left.

• In the alkaline medium, the ionisation of phenolphthalein increases considerably due to the constant removal of $\mathrm{H}^{+}$ions released from HPh by the $\mathrm{OH}^{-}$ions from the alkali. So the concentration of $\mathrm{Ph}^{-}$ion increases in the solution, which imparts pink colour to the solution. $$ \begin{array}{lll} \mathrm{HPh} & \rightleftarrows \mathrm{H}^{+}+\mathrm{Ph}^{-} \ \mathrm{NaOH} & \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \ \mathrm{H}^{+}+\mathrm{OH}^{-} & \longrightarrow \mathrm{H}_2 \mathrm{O} \end{array} $$

• For a weak acid vs strong alkali titration, phenolphthalein is the most suitable indicator. This is so because the last drop of added alkali brings the $\mathrm{pH}$ of the solution in the range in which phenolphthalein shows sharp colour change.

Methyl orange

It is a weak base and is yellow in colour in the unionised form.

Choice of Indicator

• In the titration of strong acid and a weak base, methyl orange is chosen as indicator.
• When titration between strong base and weak acid then phenolphthalein is a good indicator.
• In the titration of strong acid versus strong base either Phenolpthalein or Methyl orange can be used as indicator
• For the titration of weak acid vs weak base no indicator is available.