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 s…

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asked May 19, 2017 143k views

2 Answers

Carljohan · Answer 1 · 2017-05-19T21:42:20+0000

Answer:
M

Step-by-step 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:

$k_a=(c\alpha* c\alpha)/(c(1-\alpha))$

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

$k_a=(c^2\alpha^2)/(c)=c\alpha^2$

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

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

$1.8* 10^(-4)=0.1\alpha^2}$

$\alpha=4.2* 10^(-2)$

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

Pmaniyan · Answer 2 · 2017-05-25T19:38:31+0000

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