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Q.

What is the $[{HCOO}^{-}]$ in the solution that contains 0.015 M HCOOH and 0.02 M HCI?
$K_{a} (HCOOH) = 1.8 x 10^{- 4}$

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a

$1.35 \times 10^{- 4}$

b

$8 \times 10^{- 3}$

c

$1.8 \times 10^{- 4}$

d

$1.8 \times 10^{- 2}$

answer is B.

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Detailed Solution

$HCOOH ⇌ H^{+} + {HCOO}^{-}$

$\underset{\underset{0}{0.02}}{HCl} ⟶ {\underset{\underset{0.02}{0}}{H}}^{+} + \underset{\underset{0.02}{0}}{{Cl}^{-}}$

$K_{a} = \frac{[H^{+}] [{HCOO}^{-}]}{[HCOOH]}$

$1.8 \times 10^{- 4} = \frac{0.02 \times [{HCOO}^{-}]}{0.015}$

$[{HCOO}^{-}] = 1.35 \times 10^{- 4} M$

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What is the [HCOO-]in the solution that contains 0.015 M HCOOH and 0.02 M HCI? Ka (HCOOH)=1.8 x 10-4