0.1M NaOH is titrated with 0.1M HCOOH. The change in pH between 15 and 45 stages of neutralization of HCOOH is given by xlogy. The value of x+y is-----

α−α5=4α5pH1=pKa+logα54α5α−4α5α5pH2=pKa+log4α5α5pH2−pH1=2log4

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$0.1 M N a O H$ is titrated with $0.1 M H C O O H .$ The change in $p H$ between $\frac{1}{5}$ and $\frac{4}{5}$ stages of neutralization of $H C O O H$ is given by $x \log y .$ The value of $x + y$ is-----

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answer is 6.

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Detailed Solution

$α - \frac{α}{5} = \frac{4 α}{5}$

$\begin{array}{l} p H_{1} = p K_{a} + \log \frac{\frac{α}{5}}{\frac{4 α}{5}} \\ α - \frac{4 α}{5} \\ \frac{α}{5} \end{array}$

$\begin{array}{l} p H_{2} = p K_{a} + \log \frac{\frac{4 α}{5}}{\frac{α}{5}} \\ p H_{2} - p H_{1} = 2 \log 4 \end{array}$

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