This is a buffer mixture containing a weak base and its salt. Hence the equation to be used is
\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
pK_b = logK_b
= -log 1.8 - log 10^-5 = 4.7447
∴ \(pOH=4.7447+\log\frac { 0.15 }{ 0.10 } \)
pOH = 4.7447 + log 1.5
= 4.7447 + 0.1761 = 4.9208
pH + POH = 14
pH + 4.9208 = 14
pH = 9.08

Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }][{ H }_{ 2 }O] } \)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }] } \)  ...(1)
\({ CH }_{ 3 }{ COONH }_{ (aq) }\rightleftharpoons { C }{ H }_{ 3 }COO_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }CO{ O }^{ - }][{ H }^{ + }] }{ [{ CH }_{ 3 }{ COO }H] } \)  ...(2)
(1) x (2)
⇒ K_b . K_a = [H⁺][OH^-]
we know that [H⁺] [OH^-] = K_w
K_h· K_a = K_w
K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of Ostwald's dilution law. K_h = h²C. and i.e [OH^-] = \(\sqrt { { K }_{ h }.C } \)
pH of salt solution in terms of K_a and the concentration of the electrolyte
pH + pOH = 14
pH = 14 - pOH = 14 - {-log [OH^-]}
= 14 + log [OH^-]
∴ pH = 14 + log (K_hC)^\(\frac12\)
pH =14 + log \({ \left( \frac { { K }_{ w }C }{ { K }_{ a } } \right) }^{ \frac { 1 }{ 2 } }\)
pH = 14 +  (\(\frac12\) log K_w + \(\frac12\) log C - \(\frac12\) log K_a)
[∴ K_w = 10^-14]
\(pH=14-7+\frac { 1 }{ 2 } \log \ C+\frac { 1 }{ 2 } p{ K }_{ a }\frac { 1 }{ 2 } \log{ K }_{ w }=\frac { 1 }{ 2 } \times { \log10 }^{ -14 }=\frac { -14 }{ 2 } (1)=-7\)
\(pH=7+\frac { 1 }{ 2 } { pK }_{ a }+\frac { 1 }{ 2 } \log \ C\)   [-log K_a = pK_a]

Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and water.
\(\mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{HNO}_{3(\mathrm{aq})} \longrightarrow \mathrm{NaNO}_{3(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)
The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃^- ions.
\( \mathrm{NaNO}_{3(\mathrm{aq})} \longrightarrow \mathrm{NO}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{3(\mathrm{aq})}^{-} \)
Water dissociates to a small extent as
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
Since [H⁺] = [OH^-], water is neutral
\({ NO }_{ 3 }^{ - }\) ion is the conjugate base of the strong acid HNO₃ and hence it has no tendency to react with H⁺.
Similarly, Na⁺ is the conjugate acid of the strong base NaOH and it has no tendency to react with OH^-.
It means that there is no hydrolysis. In such cases [H⁺] = [OH^-] pH is maintained and, therefore, the solution is neutral.

Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa.
The dissociation of the buffer components occurs as below.
\( \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{sq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Na}_{(\mathrm{aq})}^{+}\)
If an acid is added to this mixture, it will be consumed by the conjugate baseCH₃COO^- to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.
\({ { CH }_{ 3 }COO }_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\rightarrow { CH }_{ 3 }{ COOH }_{ (aq) }\)
If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.

Degree if dissociation \(\alpha =\sqrt { \frac { K_{ a } }{ C } } ,C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pOH=\log\frac { 1 }{ \sqrt { { K }_{ b }.C } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75\times { 10 }^{ -5 }\times .1 } } \)
\(=\log\frac { 1 }{ \sqrt { 1.75\times { 10 }^{ -6 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.75 } } \)
\(=3-\frac { 1 }{ 2 } \log1.75\)
\(=3-\frac { 1 }{ 2 } \times 0.2430=2.8785\)
pH = 14 - pOH
∴ pH = 14 - 2.8785 = 11.1215
Alternating,
\(\left[ { H }^{ - } \right] =\sqrt { { K }_{ b }\times C } \)
\(=\sqrt { 1.75\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.75\times { { 10 }^{ -6 } } } \)
= 1.322 x 10^-3
∴ pOH = - log [H⁺] = - log (1.323 x 10^-3)
3 - 0.1216 = 2.8784
∴ pH = 14 - pOH = 14 - 2.8784 = 11.1216