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Standard XII

Chemistry

Ostwald Dilution Law

What is the p...

Question

What is the pH of solution made by adding 3.9 g $N a N H_{2}$ into water to make a 500 mL solution $K_{a (N H_{3})}$ = $2 \times 10^{- 5}$ [Na = 23, N = 14, H = 1].

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Solution

The correct option is C 9
According to given data, number of moles of $N a N H_{2}$
$[N a N H_{2}] = \frac{3.9}{39} \times \frac{1000}{500} = 0.2 M$

So pH of a solution of strong base-weak acid salt
$p H = 7 + \frac{1}{2} \times p K_{a} + \frac{1}{2} \times l o g C$

By putting given values,
$p H = 7 + \frac{1}{2} (5 - l o g 2) + \frac{1}{2} l o g 0.2 = 9$

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What is the pH of solution made by adding 3.9 g NaNH2 into water to make a 500 mL solution Ka(NH3) = 2×10−5 [Na = 23, N = 14, H = 1].

The correct option is C 9According to given data, number of moles of NaNH2[NaNH2]=3.939×1000500=0.2MSo pH of a solution of strong base-weak acid saltpH=7+12×pKa+12×log CBy putting given values,pH=7+12(5−log 2)+12log 0.2=9

What is the pH of solution made by adding 3.9 g $N a N H_{2}$ into water to make a 500 mL solution $K_{a (N H_{3})}$ = $2 \times 10^{- 5}$ [Na = 23, N = 14, H = 1].