Đáp án:
$pH = 8,47$
Giải thích các bước giải:
PT điện li: $CH_3COONa ⇄ CH_3COO^- + Na^+\\ \hspace{0,5cm}0,15\to \hspace{1,8cm}0,15$
Ta có: $K_a(CH_3COOH) = 1,75.10^{-5}$
$⇒K_b(CH_3COO^-) = \dfrac{10^{-14}}{K_a}= 5,71.10^{-10}$
$\hspace{1,2cm}CH_3COO^- + H_2O ⇄ CH_3COOH + OH^-\\BĐ:\hspace{0,5cm}0,015\\PƯ:\hspace{0,7cm}x\hspace{4,5cm}x\hspace{2cm}x\\CB:0,015-x\hspace{3,4cm}x\hspace{2cm}x$
${{\rm{K}}_{\rm{b}}}{\rm{(C}}{{\rm{H}}_{\rm{3}}}{\rm{COO ^- ) = }}{{{\rm{[}}C{H_3}COOH].{\rm{[}}O{H^ - }{\rm{]}}} \over {{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}{\rm{]}}}} = {{x.x} \over {0,015 - x}} = 5,{71.10^{ - 10}}$
$→ x=2,93.10^{-6}$
Mà: $[H^+].[OH^-]= 10^{-14}$
$→ [H^+]=3,51.10^{-9}$
$⇒pH = -lg[H^+]=8,47$
