Question

What is the OH concentration of a 0.08M solution of CH3COONa. [Ka(CH3COOH)=1.8×10^5)

Answer 1

$pH=7+ \frac{1}{2} pK a ​ + \frac{1}{2} ​ logC$

$pH=7+ \frac{1}{2} ​ (5−log(1.8))+ \frac{1}{2} ​ log(0.08)$

$pOH=14−(7+ \frac{1}{2} ​ (5−log(1.8))+ \frac{1}{2} ​ log(0.08))=5−log(0.669)$

$[OH ^ − ]=0.669×10 ^{ −5} M$