Consider a strong acid, $H_{2} S O_{4}$ in a solution having concentration of $10^{- 2} M$ . The $p H$ of the given solution of $H_{2} S O_{4}$ is :

3.7

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5.7

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1.7

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2.7

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Solution

The correct option is C 1.7
Given, $[H_{2} S O_{4}]_{i n i t i a l} = 10^{- 2} M = C$
Since it is a strong acid, it ionizes completely.
Then,

$H_{2} S O_{4} (a q .) ⇌ 2 H^{+} (a q .) + S O_{4}^{2 -} (a q .) I n i t i a l : C 00 E q u i l i b r i u m : 02 C C$

$[H^{+}]$ $= 2 C = 2 \times 10^{- 2} M$

Since, $p H = - l o g_{10} [H^{+}]$

$p H = - l o g_{10} (2 \times 10^{- 2}) = 2 - l o g_{10} 2 = 2 - 0.3 = 1.7$

Hence, $p H$ of $10^{- 2} M H_{2} S O_{4}$ is 1.7

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