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Chemistry-

General

Easy

Question

$K subscript a end subscript$ forHCN is $5 cross times 10 to the power of negative 10 end exponent$ at $25 ℃$ . For maintaining a constant $p H equals 9$ , the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is

2 mL
4 mL
8.2 mL
6.4 mL

The correct answer is: 2 mL

$p H equals negative log invisible function application open parentheses 5 cross times 10 to the power of negative 10 end exponent close parentheses plus log invisible function application open square brackets fraction numerator 5 cross times V over denominator V plus 10 end fraction divided by fraction numerator 10 cross times 2 over denominator V plus 10 end fraction close square brackets$
$9 equals negative log invisible function application open parentheses 5 cross times 10 to the power of negative 10 end exponent close parentheses plus log invisible function application fraction numerator V over denominator 4 end fraction$
On solving, $V equals 2 blank m L$

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