Asked by Lillie Tucker
Given silver oxalate (Ag2CO4 Ksp=5.40x10^-12) and silver phosphate (Ag3PO4 Ksp=8.89x10^-17)which substance has a greater concentration of the Ag+ ion? Justify your answer with calculation
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Answered by
GPT-5 mini
AI
Write the solubilities as s (mol L^-1) for the salts dissolving in pure water:
Ag2C2O4 ⇌ 2 Ag+ + C2O4^2-
Ksp = [Ag+]^2 [C2O4^2-] = (2s)^2 (s) = 4 s^3
s = (Ksp/4)^(1/3) = (5.40×10^-12 / 4)^(1/3) = (1.35×10^-12)^(1/3) ≈ 1.11×10^-4 M
[Ag+] = 2s ≈ 2.22×10^-4 M
Ag3PO4 ⇌ 3 Ag+ + PO4^3-
Ksp = [Ag+]^3 [PO4^3-] = (3s)^3 (s) = 27 s^4
s = (Ksp/27)^(1/4) = (8.89×10^-17 / 27)^(1/4) ≈ 4.26×10^-5 M
[Ag+] = 3s ≈ 1.28×10^-4 M
Conclusion: Silver oxalate (Ag2C2O4) gives a higher Ag+ concentration (≈2.22×10^-4 M) than silver phosphate (Ag3PO4, ≈1.28×10^-4 M).
Ag2C2O4 ⇌ 2 Ag+ + C2O4^2-
Ksp = [Ag+]^2 [C2O4^2-] = (2s)^2 (s) = 4 s^3
s = (Ksp/4)^(1/3) = (5.40×10^-12 / 4)^(1/3) = (1.35×10^-12)^(1/3) ≈ 1.11×10^-4 M
[Ag+] = 2s ≈ 2.22×10^-4 M
Ag3PO4 ⇌ 3 Ag+ + PO4^3-
Ksp = [Ag+]^3 [PO4^3-] = (3s)^3 (s) = 27 s^4
s = (Ksp/27)^(1/4) = (8.89×10^-17 / 27)^(1/4) ≈ 4.26×10^-5 M
[Ag+] = 3s ≈ 1.28×10^-4 M
Conclusion: Silver oxalate (Ag2C2O4) gives a higher Ag+ concentration (≈2.22×10^-4 M) than silver phosphate (Ag3PO4, ≈1.28×10^-4 M).
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