Ionic EquilibriumHard
Question
Solubility product of AgBr at certain temperature is 2.5 × 10-13. Find out solubility of AgBr in grams per litre at this temperature (Ag = 108, Br = 80)
Options
A.2.5 × 10-7
B.5 × 10-7
C.2.5 × 10-13
D.9.4 × 10-5
Solution
Ksp = S2
S =
= 5 × 10-7 mole/litre
AgBr in grams = 5 × 10-7 × 188 = 9.4 × 10-5
S =
= 5 × 10-7 mole/litreAgBr in grams = 5 × 10-7 × 188 = 9.4 × 10-5
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