Calculate $p H$ of a resultant solution of $25 m L$ of $0.1 M H C l$ , $50 m L$ of $0.01 M H N O_{3}$ and $25 m L$ of $0.1 M N a O H$ .(Round off answer to integer)

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Solution

$25 m L$ of $0.1 M H C l$ will completely neutralize $25 m L$ of $0.1 M N a O H$ .
The total volume of the solution will be $25 m L + 50 m L + 25 m L = 100 m L$ .
Thus, $50 m L$ of $0.01 M H N O_{3}$ is diluted to 100 mL.
Hence, its concentration will be reduced to one half.
Hence, the hydrogen ion concentration will b $\frac{0.01}{2} = 0.005 M$ .
The expression for the pH of the solution will be $p H = - l o g [H^{+}] = - l o g 0.005 = 2.3$ .
Hence, the pH of the resulting solution is 2.3.
Therefore answer is $2$ .

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