Question

what is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?

Answer 1

Solution:

$CH_{3}COOH=CH_{3}COO^{-}+H^{+}$

$K_{a}=1.75*10^{-5}$ so this equilibrium lies mostly to the left and there is no need for ICE tables. Under these conditions the equilibrium expression simplifies to 

$[H^{+}]= \sqrt{\smash[b]{K_{a}*C_{M}}}$

$[H^{+}]= \sqrt{\smash[b]{(1.75*10^{-5})*(10^{-3})}}$

$[H^{+}]=$ $1.32*10^{-4} M$

$pH=-log(1.32*10^{-4})$

$pH=3.88$

$pOH=14-pH=10.12$