Calculate the amount of NH42SO4 in grams which must be added to 500 ml of 0.2 M NH3 to yield a solution of pH = 9, Kb   for  NH3 = 2 × 10-5

# Calculate the amount of ${\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}$ in grams which must be added to 500 ml of 0.2 M ${\text{NH}}_{3}$ to yield a solution of pH = 9,

1. A

32.48 g

2. B

42.48 g

3. C

13.20 g

4. D

62.48 g

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### Solution:

Let ‘a’ millimoles of ${\text{NH}}_{4}^{+}$ is added to a solution

millimoles of

Given

Moles of ${\left({\text{NH}}_{4}\right)}_{2}{\text{SO}}_{4}$ added =

g

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