At $25^{\circ} C$ , the ionization constants of $C H_{3} C O O H$ and $N H_{3}$ are almost equal. The pH of a solution of $0.01 M C H_{3} C O O H$ is 4.0 at $25^{\circ} C$ . The pOH of 0.01 $M N H_{4} O H$ at the same temperature will be :

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pH of 0.01 M H2SO4 at 250 C is equal to:

The correct option is B 1.7H2SO4⇌2H++SO2−4 Since 1 mole of H2SO4 releases 2 moles of H+ 0.01 Molar H2SO4 releases 0.02 M H+ pH=−log [H+]pH=−log [0.02]pH=1.7

$p H$ of $0.01$ M $H_{2} S O_{4}$ at $25^{0}$ C is equal to:

The correct option is B $1.7$
$H_{2} S O_{4} ⇌ 2 H^{+} + S O_{4}^{2 -}$
Since 1 mole of $H_{2} S O_{4}$ releases 2 moles of $H^{+}$
0.01 Molar $H_{2} S O_{4}$ releases 0.02 M $H^{+}$
$p H = - l o g [H^{+}]$
$p H = - l o g [0.02]$
$p H = 1.7$