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Question: What is the \( pH \) of \( 0.20{\text{ }}M \) solution of formic acid? \( pK_a = 3.75 \)...

What is the pHpH of 0.20 M0.20{\text{ }}M solution of formic acid? pKa=3.75pK_a = 3.75

Explanation

Solution

the formic acid is a weak acid. It will react with water to give a strong acid and a weak base. The reaction of the formic acid is: HCOOH(aq)+H2O(l)⇌HCOO−(aq)+H3O+(aq)HCOO{H_{\left( {aq} \right)}} + {H_2}{O_{\left( l \right)}} \rightleftharpoons HCO{O^ - }_{\left( {aq} \right)} + {H_3}{O^ + }_{\left( {aq} \right)} . You can find the acid dissociation constant first using the formula: Ka=10−pKa{K_a} = {10^{ - pK_a}} . Then, after that find the pH of the solution.

Complete answer:
Formic acid is a weak acid. It does not dissociate completely in aqueous solution, and so there will be an equilibrium established between the unionized and ionized forms of the acid. Here the formic acid will undergo a reaction with water. The reaction is as follows:
HCOOH(aq)+H2O(l)⇌HCOO−(aq)+H3O+(aq)HCOO{H_{\left( {aq} \right)}} + {H_2}{O_{\left( l \right)}} \rightleftharpoons HCO{O^ - }_{\left( {aq} \right)} + {H_3}{O^ + }_{\left( {aq} \right)}
Now let the initial concentration of formic acid be 0.20 M0.20{\text{ }}M , change in the concentration be xx and at equilibrium let the concentration be 0.20−x0.20 - x
 HCOOH(aq)+H2O(l)⇌HCOO−(aq)+H3O+(aq) initial 0.20 0 0 change (−x) (+x) (+x) equilibrium (0.20−x) x x  {\text{ }}HCOO{H_{\left( {aq} \right)}} + {H_2}{O_{\left( l \right)}} \rightleftharpoons HCO{O^ - }_{\left( {aq} \right)} + {H_3}{O^ + }_{\left( {aq} \right)} \\\ initial{\text{ 0}}{\text{.20 0 0}} \\\ change{\text{ }}\left( { - x} \right){\text{ }}\left( { + x} \right){\text{ }}\left( { + x} \right) \\\ equilibrium{\text{ }}\left( {0.20 - x} \right){\text{ }}x{\text{ }}x \\\
We will first find the acid dissociation constant, Ka{K_a} , using pKapK_a
It is known that Ka=10−pKa{K_a} = {10^{ - pK_a}}
Thus Ka=10−3.75{K_a} = {10^{ - 3.75}}
⇒1.78×10−4\Rightarrow 1.78 \times {10^{ - 4}}
Therefore we got the acid dissociation constant value as 1.78×10−41.78 \times {10^{ - 4}} . This will be used in our further calculations.
For the above reaction the equilibrium constant will be written as:
Ka=[HCOO−]×[H3O+][HCOOH]K_a = \dfrac{{[HCO{O^ - }] \times [{H_3}{O^ + }]}}{{[HCOOH]}}
⇒Ka=[x]×[x][0.20−x]\Rightarrow K_a = \dfrac{{[x] \times [x]}}{{[0.20 - x]}}
Since KaK_a is small compared with the initial concentration of the acid, we can approximate (0.20−x)  (0.20 - x)\; with 0.200.20 . This will give us
Ka=x20.2K_a = \dfrac{{{x^2}}}{{0.2}}
⇒x=0.2×1.78×10−4\Rightarrow x = \sqrt {0.2 \times 1.78 \times {{10}^{ - 4}}}
⇒5.97×10−3\Rightarrow 5.97 \times {10^{ - 3}}
This is the value of xx , thus the concentration of the hydronium ion will be
x=[H3O+]=5.97×10−3Mx = [{H_3}{O^ + }] = 5.97 \times {10^{ - 3}}M
Hence the pH of the solution will be:
pHsol=−log([H3O+])p{H_{sol}} = - log([{H_3}{O^ + }])
⇒pHsol=−log(5.97×10−3)\Rightarrow p{H_{sol}} = - log(5.97 \times {10^{ - 3}})
⇒pHsol=2.22\Rightarrow p{H_{sol}} = 2.22
Therefore the pHpH of 0.20 M0.20{\text{ }}M solution of formic acid is 2.222.22 .

Note:
The above reaction is an acid-base reaction. The acid dissociation constant, KaK_a , (also known as acidity constant, or acid-ionization constant) is a quantitative measure of the strength of an acid in solution. The general equation for the dissociation constant is:
Ka=[A−][H+][HA]K_a = \dfrac{{[{A^ - }][{H^ + }]}}{{[HA]}} , where quantities in square brackets represent the concentrations of the species at equilibrium.