For 0.50 M aqueous solution of sodium cyanide, (pKb of CN−is4.70), calculate
(i) Hydrolysis constant
(ii) Degree of hydrolysis
iii) pH
log( 3.15 )=0.498 Antilog (-9.3)=5.01×10−10
A
2×105,6.3×10−3,11.5
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B
2×103,6.3×10−3,11.5
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C
4×103,6.3×10−3,9.5
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D
2×103,4.2×10−3,11.5
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Solution
The correct option is A2×105,6.3×10−3,11.5 Salt hydrolysis of NaCN is NaCN+H2O→HCN+Na++OH−
Given, pKb=4.70
For conjugate acid-base pair, pKa+pKb=14pKa=14−4.70pKa=9.3pKa=−logKa9.3=−logKaKa=5.01×10−10Kh=KwKaKh=1×10−145.01×10−10Kh=2×10−5
(ii) Degree of hydrolysis, h=√(Kh/c)=√(2×10−5/0.5)