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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)
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
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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