A 30.0 mL sample of 0.50 M $H C l O_{4}$ is titrated with a 0.25 M $K O H$ solution. The ${H_{3} O}^{+}$ concentration after the addition of 5.0 mL of $K O H$ is __________ M.

0.00125

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0.0138

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0.015

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0.393

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Solution

The correct option is D 0.393
Moles of $H C l_{4} = 0.03 \times 0.50 = 0.015$
Moles of $K O H = 0.005 \times 0.25 = 0.00125$
Moles of $H_{3} O^{+}$ or $H^{+}$ in excess $= 0.015 - 0.00125 = 0.01375$
Total volume $= 30.0 + 5.0 = 35 m L = 0.035 L$
$[H_{3} O^{+}]$ or $H^{+} = \frac{0.01375}{0.035} = 0.392$

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