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Q.

1.0 L of an aqueous solution in which $[H_{2} {CO}_{3}] = [{HCO}_{3}^{-}] = 0.10 M$ , has $[H^{+}] = 4.2 \times 10^{- 7}$ . What is the $[H^{+}]$ after 0.005 mole of $NaOH$ has been added?

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a

$\begin{aligned} 2.1 \times 10^{- 9} M \end{aligned}$

b

$\begin{aligned} 2.2 \times 10^{- 8} M \end{aligned}$

c

$3.8 \times 10^{- 7} M$

d

$4.6 \times 10^{- 7} M$

answer is C.

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Detailed Solution

$K_{a} = [H^{+}] = 4.2 \times 10^{- 7}; {[H_{2} {CO}_{3}]}_{new} = 0.1 - 0.005$

${[{HCO}_{3}^{-}]}_{new} = 0.1 + 0.005$

Now,

$4.2 \times 10^{- 7} = [H^{+}] \times \frac{0.105}{0.095}$

$\Rightarrow [H^{+}] = 3.8 \times 10^{- 7} M$ .

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