pH=-log[H⁺]
For weak acids,
\(= \sqrt{k_a \times C}\)
=\(\sqrt{1.8\times10^{-5}\times0.1}\)
=\(1.34 \times10^{-3}\) M
\(pH=-\log(1.34\times10^{-3})\)
= 3-log1.34
= 3-0.1271
= 2.8729 \(\simeq\) 2.87

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

(i) \(\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\left(25^{\circ} \mathrm{C}\right)\)
(ii) Ionic product of water is defined as the product of the concentration of hydronium and hydroxide ions of pure water. Its value at 25^oC is \(1 \times 10^{-14} \mathrm{~mol}^{2} \mathrm{dm}^{-2}\)

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.

\(pH=pK_{a}+\log[\frac{salt}{acid}]\)
Given that K_a = \(1.8\times10^{-5}\)
\(\therefore pK_{a}=-\log(1.8\times10^{-5})\)
= 5 - log 1.8
= 5 - 0.26
= 4.74
\(\therefore pH=4.74+\log\frac{0.20}{0.18}\)
= 4.74 + log(10/9)
= 4.74 + log10 - log9
= 4.74 + 1 - 0.95
= 5.74 - 0.95
= 4.79

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

\(h=\sqrt{K_{h}}=\sqrt{\frac{K_{w}}{K_{a}K_{b}}}=\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times{1.8\times10^{-5}}}}\)
\(= \frac{1 \times10^{-7}}{1.8\times10^{-5}}\)
=\(0.7453\times10^{-2}\)
\(pH=\frac{1}{2}pK_{w}+\frac{1}{2}pK_{a}-\frac{1}{2}pK_{b}\)
Given that \(K_{a}=K_{b}=1.8\times10^{-5}\)
if K_a = K_b, then, pK_a = pK_b 
\(\therefore pH= \frac{1}{2}pK_{w}=\frac{1}{2}(14)=7\)
pH = 7

HCN is a weak acid
\({\left[\mathrm{H}^{+}\right] } =\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}  \)
\(=\sqrt{10^{-9} \times 0.4}  \)
\(=\sqrt{4 \times 10^{-10}}   \)
\(=2 \times 10^{-5} \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10}\left(2 \times 10^{-5}\right)  \)
\(=-\left[\log _{10} 2-5 \log 10\right]  \)
\(=5-\log 2\)
= 5-0.3010 = 4.6990

\(\mathrm{M} =\frac{\mathrm{V}_{1} \mathrm{M}_{1}-\mathrm{V}_{2} \mathrm{M}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}}  \)
\(=\frac{(50 \times 0.05)-(50 \times 0.025)}{50+50}  \)
\(\text { Molarity }=\frac{\text { Number of millimoles }}{\mathrm{V}_{\mathrm{m} l}}\)
\(=\frac{2.5-1.25}{100}=\frac{1.25}{100}=0.0125 \mathrm{M}  \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10} 0.0125  \)
Normality = Molarity \(\times\) basicity
\(=-\log _{10}\left(1.25 \times 10^{-2}\right)  \)
\(=-\left[\log _{10} 1.25-2 \log _{10} 10\right]  \)
\(=2-\log _{10} 1.25=2-0.0969 \)
pH = 1.9031

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48