Ionic EquilibriumHard

Question

What is the minimum mass of NaBr which should be added in 200 ml of 0.0004 M-AgNO3 solution just to start the precipitation of AgBr? The value of Ksp of AgBr = 4 × 10−13 (Br = 80).

Options

A.1.0 × 10−9 g
B.2 × 10−10 g
C.2.06 × 10−8 g
D.1.03 × 10−7 g

Solution

AgBr (s) $\rightleftharpoons$ Ag+ + Br–

For precipitation, Q > Ksp

or, 0.0004 × [Br–] > 4 × 10–13

∴ [Br–] > 10–9 M

Hence, minimum mass of NaBr needed $= \left( \frac{200 \times 10^{- 9}}{1000} \right) \times 103 = 2.06 \times 10^{- 8}\text{ gm}$

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