The solubility product of solid silver(I) bromide is 7.7 x 10−13 in water at 298 K. What is its solubility in grams per liter?
A. 1.7 x 10−4 g/L
B. 8.8 x 10−7 g/L
C. 1.8 x 10−7 g/L
D.1.7 x 10−5 g/L
E. 2.2 x 10−20 g/L
Let s be the solubility of AgBr.
Then, Ksp = [Ag+][Br-]
7.7 *10^(-13) = s*s
s = 8.775*10^(-7) M
Molar mass of AgBr = 187.77 g/mol
Then, s = 8.775*10^(-7)*187.77
s = 1.65*10^(-4) = 1.7*10^(-4) g/L.
Therefore, option (A) is the correct answer.
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