Question

The concentration of formate ion $\left(HCO{O}^{-}\right)$ present in 0.1 M formic acid $\left(HCOOH\right)$ solution at equilibrium is :
$(K_a=1.6\times {10}^{-4})$

• A. $4\times {10}^{-3}M$
• B. $3\times {10}^{-6}M$
• C. $1.5\times {10}^{-3}M$
• D. $5\times {10}^{-6}M$

Answer 1

The correct option is A $4\times {10}^{-3}M$

$HCOOH\left(l\right)\rightleftharpoons HCO{O}^{-}(aq.)+H^{+}(aq.)Initial:C00Equilibrium:C(1-\alpha )C\alpha C\alpha$
At equilibrium,
$\left[HCO{O}^{-}\right]=\left[H^{+}\right]$

$K_a=\frac{\left[HCO{O}^{-}\right]\left[H^{+}\right]}{\left[HCOOH\right]}=\frac{(C\alpha {)}^2}{C(1-\alpha )}$
Since formic acid is a weak acid
$\alpha <<1.\therefore 1-\alpha \approx 1$

$1.6\times {10}^{-4}=C{\alpha }^2=0.1{\alpha }^2\alpha =\sqrt{16\times {10}^{-4}}=4\times {10}^{-2}$
Concentration of $HCO{O}^{-}$ ion at equilibrium= $C\alpha =0.1\times 4\times {10}^{-2}=4\times {10}^{-3}M$