Question

The pH of a solution obtained by mixing $100{\text{ ml}}$ of $0.2{\text{ M C}}{{\text{H}}_3}{\text{COOH}}$ with $100{\text{ ml}}$ of $0.2{\text{ M NaOH}}$ will be (${\text{p}}{{\text{K}}_{\text{a}}}$ for ${\text{C}}{{\text{H}}_3}{\text{COOH}} = 4.74$ and $\log 2 = 0.301$)
A) 4.74
B) 8.87
C) 9.10
D) 8.57

Answer 1

We know that the degree of alkalinity or acidity of a solution is known as its pH. pH is the negative logarithm of the hydrogen ion concentration. We are given acetic acid which is a weak acid and sodium hydroxide which is a strong base. To solve this we have to use the equation for the pH of weak acid and strong base.

Complete solution:
We know that the degree of alkalinity or acidity of a solution is known as its pH. pH is the negative logarithm of the hydrogen ion concentration.
We are given that $100{\text{ ml}}$ of $0.2{\text{ M C}}{{\text{H}}_3}{\text{COOH}}$ is mixed with $100{\text{ ml}}$ of $0.2{\text{ M NaOH}}$. Here, ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ is known as acetic acid which is a weak acid and ${\text{NaOH}}$ is known as sodium hydroxide which is a strong base. The reaction is as follows:
${\text{C}}{{\text{H}}_3}{\text{COOH}} + {\text{NaOH}} \to {\text{C}}{{\text{H}}_3}{\text{COONa}} + {{\text{H}}_{\text{2}}}{\text{O}}$
Thus, the salt of weak acid and strong base is formed.
First we will calculate the number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ and ${\text{NaOH}}$ in $100{\text{ ml}}$ of each.
We are given $100{\text{ ml}}$ of $0.2{\text{ M C}}{{\text{H}}_3}{\text{COOH}}$. $0.2{\text{ M C}}{{\text{H}}_3}{\text{COOH}}$ means that $0.2{\text{ mol}}$ of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ are present in $1000{\text{ ml}}$. Thus,
Number of moles of ${\text{C}}{{\text{H}}_3}{\text{COOH}}$ $= 100{\text{ ml}} \times \dfrac{{0.2{\text{ mol}}}}{{1000{\text{ ml}}}} = 0.02{\text{ mol}}$
We are given $100{\text{ ml}}$ of $0.2{\text{ M NaOH}}$. $0.2{\text{ M NaOH}}$ means that $0.2{\text{ mol}}$ of ${\text{NaOH}}$ are present in $1000{\text{ ml}}$. Thus,
Number of moles of ${\text{NaOH}}$ $= 100{\text{ ml}} \times \dfrac{{0.2{\text{ mol}}}}{{1000{\text{ ml}}}} = 0.02{\text{ mol}}$
Total volume $= \left( {100 + 100} \right){\text{ml}} = 200{\text{ ml}} = 200 \times {10^{ - 3}}{\text{ L}}$
From the reaction stoichiometry, the number of moles salt i.e. ${\text{C}}{{\text{H}}_3}{\text{COONa}}$ are $0.02{\text{ mol}}$.
Thus, the concentration of ${\text{C}}{{\text{H}}_3}{\text{COONa}}$ $= \dfrac{{0.02{\text{ mol}}}}{{200 \times {{10}^{ - 3}}{\text{ L}}}} = 0.1{\text{ M}}$
The equation to calculate the pH of weak acid strong base is as follows:
${\text{pH}} = 7 + \dfrac{1}{2}{\text{p}}{{\text{K}}_{\text{a}}} + \dfrac{1}{2}\log C$
Where $C$ is the concentration of the salt.
We are given that the ${\text{p}}{{\text{K}}_{\text{a}}}$ for ${\text{C}}{{\text{H}}_3}{\text{COOH}} = 4.74$. Thus,
${\text{pH}} = 7 + \dfrac{1}{2}\left( {{\text{4}}{\text{.74}}} \right) + \dfrac{1}{2}\log \left( {0.1} \right)$
${\text{pH}} = 7 + \dfrac{1}{2}\left( {{\text{4}}{\text{.74}} + \log \left( {0.1} \right)} \right)$
${\text{pH}} = 7 + \dfrac{1}{2}\left( {{\text{4}}{\text{.74}} + \left( { - 1} \right)} \right)$
${\text{pH}} = 7 + \dfrac{1}{2}\left( {3.74} \right)$
${\text{pH}} = 7 + 1.87$
${\text{pH}} = 8.87$
Thus, the pH of a solution obtained by mixing $100{\text{ ml}}$ of $0.2{\text{ M C}}{{\text{H}}_3}{\text{COOH}}$ with $100{\text{ ml}}$ of $0.2{\text{ M NaOH}}$ is 8.87.

Thus, the correct option is (B).

Note: If the pH of the solution is less than 7 then the solution is acidic in nature. If the pH of the solution is equal to 7 then the solution is neutral in nature. If the pH of the solution is more than 7 then the solution is basic in nature. Here, the pH is 8.87 thus, we can say that the solution is basic or alkaline in nature.

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