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

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syed514 syed514 · Answer 1 · 2016-11-11T16:15:26+01:00

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

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kobenhavn kobenhavn · Answer 2 · 2019-03-18T07:56:11+01:00

Answer: [tex]4.2\times 10^{-3}[/tex] M

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

[tex]HCOOH\rightleftharpoons HCOO^-+H^+[/tex]

initially conc. c 0 0

At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]

The expression for dissociation constant is:

[tex]k_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}[/tex]

when [tex]\alpha[/tex] is very very small the, the expression will be,

[tex]k_a=\frac{c^2\alpha^2}{c}=c\alpha^2[/tex]

[tex]c=\frac{moles}{\text {Volume in L}}=\frac{0.1}{1}=0.1M[/tex]

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

[tex]1.8\times 10^{-4}=0.1\alpha^2}[/tex]

[tex]\alpha=4.2\times 10^{-2}[/tex]

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