When 50 ~mL of 0.1 M NH_3 is mixed with 10 ~mL of 0.1 M HCl then what is the pH of resultant solution? (p K_b=4.75)

Correct answer: C. 9.85 aligned & NH_3+HCl NH_4 Cl & Initial 50 × 0.1 10 × 0.1 & =5 mmol =1 mmol & After 4 mmol 0 mmol 1 mmol & reaction & pOH=p K_b+ [ salt ] [ base ] & =4.75+ ((1)/(4))=4.15 & pH=14-pOH=14-4.15=9.85 & aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Chemistry
Chapter: Ionic Equilibrium

AIIMS 2018 · 26 May · Chemistry · Ionic Equilibrium

When 50 ~mL of 0.1 M NH_3 is mixed with 10 ~mL of 0.1 M HCl then...

When

50 \mathrm{~mL}

0.1 \mathrm{M} \mathrm{NH}_3

is mixed with

10 \mathrm{~mL}

0.1 \mathrm{M} \mathrm{HCl}

then what is the

\mathrm{pH}

of resultant solution?

\left(\mathrm{p} K_b=4.75\right)

Options

A $9.25$
B $10$
C $9.85$
D $4.15$

Correct answer

C. $9.85$

Step-by-step solution

\begin{aligned} & \text{ } \quad \mathrm{NH}_3+\mathrm{HCl} \longrightarrow \mathrm{NH}_4 \mathrm{Cl} \\ & \text{ Initial } 50 \times 0.1 \quad 10 \times 0.1 \\ & =5 \mathrm{mmol} \quad=1 \mathrm{mmol} \\ & \text{ After } \quad 4 \mathrm{mmol} \quad 0 \mathrm{mmol} \quad 1 \mathrm{mmol} \\ & \text{\text{ reaction } } \\ & \mathrm{pOH}=\mathrm{p} K_b+\log \frac{[\text{ salt }]}{[\text{ base }]} \\ & =4.75+\log \left(\frac{1}{4}\right)=4.15 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-4.15=9.85 \\ & \end{aligned}

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