Question

When $50mL$ of $0.1M$ ${NH}_{4}OH$ is added to $50mL$ of $0.05M$ $HCl$ solution, the pH is:

• A. $1.6021$
• B. $12.3979$
• C. $4.7447$
• D. $9.2553$

Answer 1

The correct option is D $9.2553$

Given,

$0.1M$ of $N{H}_{4}OH$ is present in $50ml$ solution.

$0.05M$ of HCl is present in $50ml$ solution.

$N{H}_{4}OH$ - weak base and HCl - strong acid

The reaction between $N{H}_{4}OH$ and HCl can be written as,

$N{H}_{4}OH+H^{+}C{l}^{-}\to N{H_4}^{+}C{l}^{-}+H_{2}O$

Since we know that strong acid 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 hydrochloric acid present in the solution=$50\times 0.05$

=$2.5$

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

=$5$

Thus, $2.5m.mol$ of HCl neutralises $2.5m.mol$ of $N{H}_{4}OH$ forming $N{H}_{4}Cl$ (Ammonium chloride) salt.

Therefore, the number of millimoles present in ammonium chloride salt= $2.5mmol$

This leads to the formation of a weak base and its salt buffer since ammonium hydroxide is not completely neutralized by hydrochloric acid and the remaining $2.5mmol$ of ammonium hydroxide is still present.

Molarity of ammonium chloride present in $100ml$ solution ($50ml$ HCl+$50ml$ $N{H}_{4}OH$)=$\frac{2.5}{100}$

=$0.025M$

It is known that,

$pOH=p{K}_b+\log \frac{\left[Salt\right]}{\left[Base\right]}$

$pOH=4.7447+\log \frac{\left[0.025\right]}{\left[0.1\right]}$

$pOH=4.7447-0.6020$

$pOH=4.1427$

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

$\Rightarrow pH=14-pOH$$\Rightarrow pH=14-pOH$

$\Rightarrow pH=14-4.1427$

$\Rightarrow pH=9.8573$

Thus, among the four option, $9.8573\sim 9.2553$)

Therefore, the approximate value for pH that we calculated is given as $9.2553$