Question

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If $pK_b$ of ammonia solution is 4.75, the pH of the mixture will be :

• A. 3.75
• B. 4.75
• C. 8.25
• D. 9.25

Answer 1

Answer: (D)

Number of millimoles of $ NH _{3} = 50 mL \times 0.2 M =10 mmol$
Number of millimoles of HCl = $25\, mL \times 0.2 M = 5 mmol$
$5 \,mmol \,NH _{3} $ will react with $ 5\, mmol\, HCl$ to form $5\, mmol\, NH _{4} Cl$
$10-5 = 5 \,mmol \,NH _{3} $will remain.
$pOH = pK _{ b }-\log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right]}$
$pOH =4.75-\log \frac{5 / V }{5 / V }$
$pOH =4.75$
$pH =14- pOH =14-4.75=9.25$