A 0.1 M sodium acetate solution was prepared. The kh=5.6×10−10

αh=khC0=5.6×10−110−1=7.48×10−5CH3COO−aq+H2Ol⇄ CH3COOH+OH−aqt=0                                 C0t=teq                   C0−C0αh                      C0αh                          C0αwOH−=C0αw=10−1×7.48×10−5=7.48×10−6∴ pOH=5.12                        pn=8.88