Question

Calculate the concentration of the formate ion present in 0.100 M formic acid (HCOOH) solution at equilibrium (${K_a}$= 1.7$ \times {10^{ - 4}}$).
a.) $4.1 \times {10^{ - 3}}$M
b.) $3.1 \times {10^{ - 3}}$M
c.) $2.1 \times {10^{ - 3}}$M
d.) $5.1 \times {10^{ - 3}}$M

Answer 1

Hint: We know the reaction of dissociation of formic acid can be written as -
$HCOOH \rightleftharpoons {H^ + } + HCO{O^ - }$
Assuming the initial concentration of formate and proton 0 and at equilibrium ‘x’; the equilibrium constant is given as -
${K_a}$= $\dfrac{{{x^2}}}{{(0.1 - x)}}$
From this, the value of ‘x’ will give the concentration of formate ion.

Complete step by step solution:
First, let us write what is given to us and what we need to find out.
Given :
Concentration of the formic acid = 0.1 M
${K_a}$= 1.7$ \times {10^{ - 4}}$
To find :
The concentration of the formate ion
We know the reaction of dissociation of formic acid can be written as -
$HCOOH \rightleftharpoons {H^ + } + HCO{O^ - }$
The initial concentration of formic acid (HCOOH) = 0.1 M
The initial concentration of protons = 0
The initial concentration of formate ions = 0
When the reaction reaches equilibrium,
The concentration of formic acid (HCOOH) = 0.1 - x
The concentration of protons = x
The concentration of formate ions = x
We know ${K_a}$= $\dfrac{{{x^2}}}{{(0.1 - x)}}$
1.7$ \times {10^{ - 4}}$= $\dfrac{{{x^2}}}{{(0.1 - x)}}$
As ‘x’ is very small; so this can be ignored.
1.7$ \times {10^{ - 4}}$= $\dfrac{{{x^2}}}{{0.1}}$
${x^2}$= 1.7$ \times {10^{ - 5}}$
${x^2}$= 1.7$ \times {10^{ - 6}}$
‘x’ = $4.1 \times {10^{ - 3}}$M

So, the option a.) is the correct answer.

Note: It must be noted that equilibrium is a state at which the rate of forward reaction is equal to the rate of backward reaction. The amount of product formed is equal to the amount of reactant that has converted back into reactant.