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)

$\Rightarrow K_h=\frac{\left[C{H}_{3}COOH\right][\stackrel{\ominus }{H}]}{\left[C{H}_{3}COO\right]}=\frac{K_w}{K_a}=\frac{c{h}^2}{(1-h)}$

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

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

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

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

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

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