Ka for HCN is 5×10−10 at 25∘C. For maintaining a constant pH of 9, the volume of 5MKCN solution required to be added to 10mL of 2MHCN solution is (log2=0.3andlog5=0.69)
A
4 mL
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B
8 mL
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C
2 mL
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D
10 mL
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Solution
The correct option is C 2 mL ka for HCN=5×10−10 volume of 5 M KCN solution required (Vmv)=? pKa=−logKa=−log(5×10−10)=10−log5=9.3 We know that pH=pKa+log[saltacid]⇒9=9.3+log[5×V10×2]⇒−0.3=log[V4]⇒0.3=log[4V]⇒2=4V⇒V=2mL