Question

# The solution of a salt of a weak acid and weak base will have $pH:({K_b} = 1 \times {10^{ - 6\,}}and\,{K_a} = 1 \times {10^{ - 4}})$A. $7.0$ B. $8.0$ C. $6.0$ D. $4.0$

Hint:Weak electrolyte (weak acid or weak base) is dissociated to small extent in solution and there is an equilibrium between the undissociated electrolyte and the ions formed in solution.
Formula used: ${P^H} = 7 + \dfrac{1}{2}[{P^{{K_a}}} - {P^{{K_b}}}]$

Given:
${K_a} = 1 \times {10^{ - 4}};\,{P^{{K_a}}} = \, - log({K_a}) = - \log ({10^{ - 4}}) = 4$
${K_b} = 1 \times {10^{ - 6}};\,{P^{{K_b}}} = \, - log({K_b}) = - \log ({10^{ - 6}}) = 6 (\because \,log10 = 1)$
Where ${K_a}$ and ${K_b}$ are the dissociation constant of acid and base
We know that
${P^H}$of the solution of a salt of a weak acid and weak base
${P^H} = 7 + \dfrac{1}{2}[{P^{{K_a}}} - {P^{{K_b}}}]$ ………….. (i) (Where ${P^{{K_a}}}$and ${{\text{P}}^{{K_b}}}$are the power of dissociation constant of weak acid and weak base)
Put the value of ${P^{{K_a}}}$and ${{\text{P}}^{{K_b}}}$ in equation (i) then we get
${P^H} = 7 + \dfrac{1}{2}[4 - 6] \\ = 7 + \dfrac{1}{2}[ - 2] \\ {P^H} = 7 - 1 \\ {P^H} = 6 \\$
Hence the correct option is C.

Note:
Salt of weak acid and weak base undergoes both cationic and anionic hydrolysis. In cationic hydrolysis it gives a weak base and in anionic hydrolysis it gives a weak acid.
The ionization constant is a measure of acid strength: the higher the Ka value, the greater the number of hydrogen ions liberated per mole of acid in solution and the stronger is the acid. $Ka,{\text{ }}pKa,{\text{ }}Kb$ and $pKb$ are most helpful when predicting whether a species will donate or accept protons at a specific pH value. They describe the degree of ionization of an acid or base.