What is the molar solubility of AgCl (Ksp = 1.80 x 10¹0) in 0.320 M NH,? (Kf of Ag(NH₂), is 1.7 x 10')

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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What is the molar solubility of AgCl (Ksp = 1.80 x 101°) in 0.320 M NH₂?
(Kf of Ag(NH₂), is 1.7 x 10')
Question 3 of 3
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Your submitted answer contains too many
significant figures.
Your submission:
1.76 x 10¹2 M
Feedback:
Consider the balanced chemical equation for this
process. Because these two equilibria are
dependent on each other, the overall equilibrium
constant for solubility equals the Ksp multiplied
by the Kf. Complex ion formation often increases
the solubility of a substance. To calculate molar
solubility, use the ICE method and substitute into
the equilibrium expression.
Transcribed Image Text:What is the molar solubility of AgCl (Ksp = 1.80 x 101°) in 0.320 M NH₂? (Kf of Ag(NH₂), is 1.7 x 10') Question 3 of 3 Incorrect, 2 attempts remaining Your submitted answer contains too many significant figures. Your submission: 1.76 x 10¹2 M Feedback: Consider the balanced chemical equation for this process. Because these two equilibria are dependent on each other, the overall equilibrium constant for solubility equals the Ksp multiplied by the Kf. Complex ion formation often increases the solubility of a substance. To calculate molar solubility, use the ICE method and substitute into the equilibrium expression.
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