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A 0.1 M sodium acetate solution was prepared. The
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By Expert Faculty of Sri Chaitanya
a
The degree of hydrolysis is 7.48×10−5
b
The OH- concentration is 7.48×10-3M
c
TheOH- concentration is 7.48×10−6M
d
The pH is approximately 8.88
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Correct option is A
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=8.88Similar Questions
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