The pH and pOH of $0.1 M$ aqueous solution of $H {N O}_{3}$ are

0.14

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14, 0

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13, 1

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1, 13

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Solution

The correct option is D $1, 13$
$H^{+}$ ion concentration in aqueous solution of $H N O_{3}$ is $0.1 M = 10^{- 1}$
$[H^{+}] = 0.1$
So $p H = - l o g [H^{+}] = - l o g [0.1] = - (- 1) = 1$
We know that $p H + p O H = 14$
$p O H = 14 - 1 = 13$
$p H, p O H$ respectively are $1, 13.$

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