Question

Formic acid (HCO2H) has a dissociation constant of 1.8 _ 104 M. The acid dissociates 1:1. What is the [H+] if 0.1 mole of formic acid is dissolved in 1.0 liter of water? Use a scientific calculator.
0.9 _ 10^-3 M

4.2 _ 10^-3 M

7.9 _ 10^-3 M

15 _ 10^-4 M

Answer 1

Use the equation for Ka => 1.8 x 10 ^ (-4) = [H+][HCO2-}, these 2 concentrations are equal so, 
1.8 x 10 ^ (-4) = [H+]^2 

and [H+] = 4.2 × 10^-3 M 

Answer 2

Answer: $4.2\times 10^{-3}$ M

Explanation: The equilibrium reaction for dissociation of weak acid is,

$HCOOH\rightleftharpoons HCOO^-+H^+$

initially conc.       c                          0         0

At eqm.   $c(1-\alpha)$        $c\alpha$     $c\alpha$

The expression for dissociation constant is:

$k_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}$

when $\alpha$ is very very small the, the expression will be,

$k_a=\frac{c^2\alpha^2}{c}=c\alpha^2$

$c=\frac{moles}{\text {Volume in L}}=\frac{0.1}{1}=0.1M$

Now put all the given values in this expression, we get

$1.8\times 10^{-4}=0.1\alpha^2}$

$\alpha=4.2\times 10^{-2}$

$[H^+]=c\alpha=0.1\\times 4.2\times 10^{-2}=4.2\times 10^{-3}M$