A 0.1 M sodium acetate solution was prepared. The kh=5.6×10−10
The degree of hydrolysis is 7.48×10−5
The OH- concentration is 7.48×10-3M
TheOH- concentration is 7.48×10−6M
The pH is approximately 8.88
αh=khC0=5.6×10−110−1=7.48×10−5
CH3COO−aq+H2Ol⇄ CH3COOH+OH−aqt=0 C0t=teq C0−C0αh C0αh C0αw
OH−=C0αw=10−1×7.48×10−5
=7.48×10−6
∴ pOH=5.12 pn=8.88