Answer to Question #77877 in General Chemistry for Malia Commerce

Question #77877

1.At 25 ℃, 0.831 mg AgBr dissolves in 9.25 L of water. What is the equilibrium constant for the reaction below?

AgBr(s) ⇆ Ag+(aq) + Br–(aq)

Expert's answer

Answer:
Ksp = [Ag+][Br-]
0.831 mg = 0.000831 g
n = m/M
0.000831 g / 187.78 g/mole = 4.425×10^-6 mole
с = n/V
c = 4.425×10^-6 / 9.25 = 4.78×10^-7 M
Ksp = (4.78×10^-7)^-2 = 2.28×10^-13

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11.11.20, 20:23

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07.11.20, 23:03

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06.02.19, 15:10

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Answer to Question #77877 in General Chemistry for Malia Commerce

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