Question

When $50mL$ of $0.1M$ $NaOH$ is added to $50mL$ of $0.05M$ ${CH}_{3}COOH$ solution. The pH of the solution is:

• A. $1.6021$
• B. $12.3979$
• C. $4.7447$
• D. $8.7218$

Answer 1

The correct option is B $12.3979$

$0.1M$ of NaOH is present in $50ml$ solution.

$0.05M$ of $C{H}_{3}COOH$ is present in $50ml$ solution.

$C{H}_{3}COOH$ - weak acid and NaOH - strong base

The reaction between $C{H}_{3}COOH$ and NaOH can be written as,

$C{H}_{3}COOH+N{a}^{+}O{H}^{-}\to C{H}_{3}CO{O}^{-}N{a}^{+}+H_{2}O$

Since we know that strong bases and salts are easily dissociated as ions.

It is known that the product of volume in milliliters and molarity gives the number of millimoles of the acid or base.

The number of millimoles of acetic acid present in the solution=$50\times 0.05$

=$2.5$

The number of millimoles of sodium hydroxide present in the solution=$50\times 0.1$

=$5$

Thus, $2.5m.mol$ of NaOH neutralises $2.5m.mol$ of $C{H}_{3}COOH$ forming $C{H}_{3}COONa$ (sodium acetate) salt.

Therefore, the number of millimoles present in sodium acetate salt= $2.5m.mol$

This leads to the formation of the weak acid and its salt buffer since acetic acid is not completely neutralized by sodium hydroxide and the remaining $2.5m.mol$ of excess sodium hydroxide is still present.

Molarity of excess sodium hydroxide present in $100ml$ solution ($50ml$ NaOH+$50ml$ $C{H}_{3}COOH$ )=$\frac{2.5}{100}$

=$0.025M$

It is known that,

$pOH=-\log \left[O{H}^{-}\right]$

$\Rightarrow pOH=-\log \left[0.025\right]$

$\Rightarrow pOH=1.6021$

It is also known that $pH+pOH=14$

$\Rightarrow pH=14-pOH$

$\Rightarrow pH=14–1.6021$

$\Rightarrow pH=12.3979$