Question

For anionic hydrolysis, pH is given by?
A.$pH\quad =\quad 1/2pK_{ w }=1/2pK_{ b }-1/2logC$
B.$pH\quad =\quad 1/2pK_{ x }+1/2pK_{ a }-1/2logC$
C.$pH\quad =\quad 1/2pK_{ w }+1/2pK_{ a }+1/2logC$
D.None of the above

Answer 1

Hint: You will get an idea from the definition of salt hydrolysis. It is defined as a reaction in which cation or anion or both of a salt react with water to produce acidity or alkalinity. Now try to find an answer accordingly for anionic hydrolysis.

Complete step by step answer:
Anionic hydrolysis - Salts of weak acids and strong bases undergo anionic hydrolysis and yield a basic solution. In anionic hydrolysis, the solution becomes slightly basic (pH >7)
Now, we will derive a general equation of pH of anionic hydrolysis.
A general hydrolysis reaction for a salt of a weak acid (HA) and strong base can be written as,
\[A^{ - }\quad \quad \quad \quad \quad +\quad \quad \quad \quad H_{ 2 }O\quad \quad \quad \quad \quad \rightleftharpoons \quad \quad \quad \quad HA\quad \quad \quad \quad \quad +\quad \quad \quad \quad OH^{ - }\]
C(1-x) Cx Cx
Thus, $OH^{ - }$ ion concentration increases, the solution be­comes alkaline.
Applying law of mass action,
${ K }_{ h }$ = [HA][$OH^{ - }$]/[$A^{ - }$] = (Cx×Cx)/C(1-x) = (${ Cx }^{ 2 }$)/(1-x) …………(i)

Other equations present in the solution are,
            \[HA\quad \rightleftharpoons \quad A^{ - }\quad +\quad H^{ + }\],
${ K }_{ a }$ = [$A^{ - }$][$H^{ + }$]/[HA] ...... (ii)
\[H_{ 2 }O\quad \rightleftharpoons \quad H^{ + }\quad +\quad OH^{ - }\],
${ K }_{ w }$ = [$H^{ + }$][$OH^{ - }$] ....... (iii)
From eqs. (ii) and (iii) we can conclude that,
$log[OH^{ - }]\quad =\quad logK_{ w }-logK_{ a }+log[salt]/[acid]$
$-pOH\quad =\quad -pK_{ w }+pK_{ a }+log[salt]/[acid]$
$pK_{ w }-pOH\quad =\quad pK_{ a }+log[salt]/[acid]$
Considering eq. (i) again we can write,
$K_{ x }=Cx^{ 2 }/(1-x)$ or $K_{ h }=Ch^{ 2 }/(1-h)$
When h is very small, (1-h) $\rightarrow$ 1
or $h^{ 2 }=K_{ h }/C$
or $h=\sqrt { K_{ h }/C }$
[$OH^{ - }$] = h × C = $h=\sqrt { K_{ h }C }$ = $h=\sqrt { C\times { K }_{ w }/{ K }_{ a } }$
$[H^{ + }]\quad =\quad K_{ w }/[OH^{ - }]\quad =\quad K_{ w }/\sqrt { (CK_{ w }/K_{ a }) } =\sqrt { (K_{ a }K_{ w })/K_{ c } } $
$-log[H^{ + }]=-1/2logK_{ w }-1/2logK_{ a }+1/2logC$
            $pH\quad =\quad 1/2pK_{ w }+1/2pK_{ a }+1/2logC$
This is the equation for pH in anionic hydrolysis.

Therefore, the correct answer to this question is C.

Note: In chemistry, pH is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH.
pH for cationic hydrolysis is given by,
$pH\quad =\quad 1/2pK_{ w }-1/2pK_{ b }-1/2logC$