Question

Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate $(K_a = 1.0 \times 10^{-5})$ will be :

• A. 5.0
• B. 6.0
• C. 8.0
• D. 9.0

Answer 1

Answer: (D)

$\Rightarrow K _{ h }=\frac{\left[ CH _{3} COOH \right][ \overset{⊖}{ H} ]}{\left[ CH _{3} COO \right]}=\frac{ K _{ w }}{ K _{ a }}=\frac{ ch ^{2}}{(1- h )} $

$\Rightarrow \frac{10^{-14}}{10^{-5}}= ch ^{2}$

$\{\because$ h is very small $\therefore 1- h \approx 1\}$

$\Rightarrow h =\sqrt{\frac{10^{-9}}{0.1}}=10^{-4}$

$\therefore [ \overset{⊖}{O}H ]= ch =0.1 \times 10^{-4}=10^{-5} $

$\Rightarrow [ \overset{\oplus}{H }]=10^{-9}$

$\therefore p ^{ H }=-\log [ \overset{\oplus}{H} ]=9$