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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

Verified Answer :

syed514

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: \(4.2\times 10^{-3}\) 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

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

\(\alpha=4.2\times 10^{-2}\)

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

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