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 :

$(a)\;5.0\qquad(b)\;6.0\qquad(c)\;8.0\qquad(d)\;9.0$

Kh=kw/ka =10^-14/10^-5 =10^-9 h=√kh/c h=√10^-9/0.1 So h becomes 10^-5 We know that concentration of acid can be written as [H+]= CH [H+] becomes 10^ -5 Therefore pH happens to be 5