Question

The approximate $pH$ of a solution formed by mixing equal volumes of solutions of $0.1\, M$ sodium propionate and $0.1\, M$ propanoic acid (the dissociation constant of propanoic acid is $1.3 \times 10^{-5} \,mol\, dm ^{-3}$ ) will be

• A. 2.45
• B. 4.89
• C. 5.98
• D. 6.89

Answer 1

Answer: (B)

As $CH _{3} CH _{2} COONa$ is a strong electrolyte.

Now consider solution of $0.1\, M$ propanoic acid having initially $0.1 \,M \,CH _{3} CH _{2} COO ^{-}$.

$Ka =\frac{\left[ CH _{3} CH _{2} COO ^{-}( aq )\right]\left[ H ^{+}\right]}{\left[ CH _{3} CH _{2} COOH \right]}=1.3 \times 10^{-5}$
$\frac{(0.1+ X )( X )}{(0.1- X )}=1.3 \times 10^{-5}$
Assuming that $0.1+X=0.1$
$0.1-X=0.1$, we get
$\frac{(0.1)( X )}{(0.1)}=1.3 \times 10^{-5}$
or $X =1.3 \times 10^{-5}=\left[ H ^{+}\right]$
So $pH =-\log \left(1.3 \times 10^{-5}\right)=4.89$