10 mL of a strong acid solution of pH = 2 is mixed with 990 mL of another strong acid solution of pH = 4. The pH of resulting solution will be :

4.002

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4.2

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3.7

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Solution

The correct option is D 3.7

$p H_{s o l u t i o n 1} = 2; M_{1} = [H^{+}] = 10^{- 2} p H_{s o l u t i o n 2} = 4; M_{2} = [H^{+}] = 10^{- 4} M_{3} = [H^{+}]_{a f t e r m i x i n g} V_{1} = 10 m L, V_{2} = 990 m L V_{3} = V_{1} + V_{2} = 1000 m L A p p l y i n g :$
$M_{1} V_{1} + M_{2} V_{2} = M_{3} V_{3} M_{3} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{3}} = \frac{(10^{- 2} \times 10) + (10^{- 4} \times 990)}{1000} = \frac{0.1 + 0.0990}{1000} = 1.99 \times 10^{- 4} p H = - l o g_{10} [1.99 \times 10^{- 4}] p H = 3.7 A n s .$

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