Question

The concentration of hydrogen ions is a 0.2M solution of formic acid is $6.4 \times {10^{ - 3}}mol{L^{ - 1}}$. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to 1 $mol{L^{ - 1}}$. What will be the pH of this solution?
The dissociation of formic acid is $2.4 \times {10^{ - 4}}$(write the value to the nearest integer)

Answer 1

Hint: The acid dissociation constant is measured by dividing the concentration of individual cation and anion formed by dissociation of acid by the concentration of acid. The pH of acid is the negative logarithm of hydrogen ion concentration where concentration of hydrogen ion should be determined first.

Complete step by step answer:
The molarity of formic acid is 0.2M.
The concentration of hydrogen ion is $6.4 \times {10^{ - 3}}mol{L^{ - 1}}$
The concentration of sodium formate is 1$mol{L^{ - 1}}$.
The dissociation of formic acid is $2.4 \times {10^{ - 4}}$
The formic acid undergoes dissociation reaction to give its constituent anion and cation.
The reaction for dissociation of formic acid is shown below.
$HCOOH \to HCO{O^ - } + {H^ + }$
In this reaction, formic acid dissociates into formate anion and hydrogen ion.
The initial concentration of formic acid is 0.2 M, formate ion is $6.4 \times {10^{ - 3}}mol{L^{ - 1}}$, hydrogen ion is $6.4 \times {10^{ - 3}}mol{L^{ - 1}}$.
The equilibrium concentration of formic acid is 0.2 + x, formate ion is 1$mol{L^{ - 1}}$, hydrogen ion is $6.4 \times {10^{ - 3}}mol{L^{ - 1}}$ - x
The dissociation constant is calculated by dividing the concentration of the individual constituent ions by the total concentration of the acid solution.
The dissociation constant of acid HA is given as shown below.
\[{K_A} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}\]
Where, ${K_A}$is a acid dissociation constant.
The dissociation constant of formic acid is shown below.
${K_A} = \dfrac{{[HCO{O^ - }][{H^ + }]}}{{[HCOOH]}}$
Substitute the values of concentration in the above equation.
$\Rightarrow {K_A} = \dfrac{{[1][6.4 \times {{10}^{ - 3}} - x]}}{{[0.2 + x]}} = 2.4 \times {10^{ - 4}}$
The value of x after solving the equation is 0.00635.
Substitute the value of x in the equilibrium concentration of ${H^ + }$.
$\Rightarrow {H^ + } = 6.4 \times {10^{ - 3}} - 0.00635$
$\Rightarrow {H^ + } = 4.9 \times {10^{ - 5}}$
The pH of an acid is defined as the negative logarithm of hydrogen ion concentration.
$pH = - \log [{H^ + }]$
Substitute the value of ${H^ + }$ in the above equation.
$\Rightarrow pH = - \log [4.9 \times {10^{ - 5}}]$
$\Rightarrow pH = 4.305$
Therefore, the nearest integer is 4.

Note:
The acid dissociation constant measures the acidic strength in the solution and is calculated at equilibrium condition. The concentration of ions is said to be in equilibrium when the concentration does not change with time.