Question
In 500 ml of 2.5 × 10−5 M – AgNO3 solution, 2000 ml of 5.0 × 10−2 M – NaCl solution is added. The mass of precipitate of AgCl formed is (Ksp of AgCl = 2 × 10−10, Ag = 108)
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Solution
$Ag^{+} + Cl^{-} \rightleftharpoons AgCl(s);K_{eq} = \frac{1}{K_{sp}} = \frac{1}{2 \times 10^{- 10}}$
$\frac{500 \times 2.5 \times 10^{- 5}}{2500}$ $\frac{2000 \times 5 \times 10^{- 2}}{2500}$
= 5 × 10–6 M = 4 × 10–2 M
100% 0 (4 × 10–2 – 5 × 10–6) ≈ 4 × 10–2 M
Equ. x M (4 × 10–2 + x) ≈ 4 × 10–2 M
Now, $\frac{1}{2 \times 10^{- 10}} = \frac{1}{x \times 4 \times 10^{- 10}} \Rightarrow x = 5 \times 10^{- 9}$
Hence, Ag+ ppt = (5 × 10–6 – 5 × 10–9) ≈ 5 × 10–6 M
∴ Mass of AgCl ppt = $\left( \frac{2500 \times 5 \times 10^{- 6}}{1000} \right) \times 143.5 = 1.79 \times 10^{- 2}\text{ gm}$
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