Answer
$Ka = 1.8\times 10^{- 4}$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.38}$
$[H_3O^+] = 4.169 \times 10^{- 3}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [HCOO^-] = x$
-$[HCOOH] = [HCOOH]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$
$Ka = \frac{x^2}{[InitialHCOOH] - x}$
$Ka = \frac{( 4.169\times 10^{- 3})^2}{ 0.1- 4.169\times 10^{- 3}}$
$Ka = \frac{ 1.738\times 10^{- 5}}{ 0.09583}$
$Ka = 1.8\times 10^{- 4}$
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