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Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF. Solution (a) Here, Ca2+ is the common ion: CaF2 (s) = Ca2+ (aq) + 2F- (aq) Initial (M) Eq (M) I 0.010 0.010+x= 0.010 Ksp = [Ca2+] [F] => Ksp = (0.010) (2x)² 0 2x => 3.9 × 10-11 = (0.010) (4 x = 3.1 x 10-5 M 2 x = 6.2 × 10-5 M = [F] 6.2 x 10-5 mol F- 1 mol CaF2 Thus: X = 3.1 x 10-5 M CaF2 2 mol F- L Thus, 3.1 x 10-5 mol CaF2 should dissolve in 1.0 L of 0.010M Ca(NO3)2

Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF. Solution (a) Here, Ca2+ is the common ion: CaF2 (s) = Ca2+ (aq) + 2F- (aq) Initial (M) Eq (M) I 0.010 0.010+x= 0.010 Ksp = [Ca2+] [F] => Ksp = (0.010) (2x)² 0 2x => 3.9 × 10-11 = (0.010) (4 x = 3.1 x 10-5 M 2 x = 6.2 × 10-5 M = [F] 6.2 x 10-5 mol F- 1 mol CaF2 Thus: X = 3.1 x 10-5 M CaF2 2 mol F- L Thus, 3.1 x 10-5 mol CaF2 should dissolve in 1.0 L of 0.010M Ca(NO3)2

Chemistry by OpenStax (2015-05-04)
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter15: Equilibria Of Other Reaction Classes
Section: Chapter Questions
Problem 16E: Assuming that no equilibria other than dissolution are involved, calculate the concentration of all...
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Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF. Solution (a) Here, Ca2+ is the common ion: CaF2 (s) = Ca2+ (aq) + 2F- (aq) Initial (M) Eq (M) I 0.010 0.010+x= 0.010 Ksp = [Ca2+] [F] => Ksp = (0.010) (2x)² 0 2x => 3.9 × 10-11 = (0.010) (4 x = 3.1 x 10-5 M 2 x = 6.2 × 10-5 M = [F] 6.2 x 10-5 mol F- 1 mol CaF2 Thus: X = 3.1 x 10-5 M CaF2 2 mol F- L Thus, 3.1 x 10-5 mol CaF2 should dissolve in 1.0 L of 0.010M Ca(NO3)2
Transcribed Image Text:Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF. Solution (a) Here, Ca2+ is the common ion: CaF2 (s) = Ca2+ (aq) + 2F- (aq) Initial (M) Eq (M) I 0.010 0.010+x= 0.010 Ksp = [Ca2+] [F] => Ksp = (0.010) (2x)² 0 2x => 3.9 × 10-11 = (0.010) (4 x = 3.1 x 10-5 M 2 x = 6.2 × 10-5 M = [F] 6.2 x 10-5 mol F- 1 mol CaF2 Thus: X = 3.1 x 10-5 M CaF2 2 mol F- L Thus, 3.1 x 10-5 mol CaF2 should dissolve in 1.0 L of 0.010M Ca(NO3)2
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