CaF 2(s) ← Ca2+(aq) + 2 F−(aq) K = 4.0 × 10−11sp49. The concentration of F (aq)− in drinking water that isconsidered to be ideal for promoting dental health is4.0 × 10−5 M. Based on the information above, the maximumconcentration of Ca 2+(aq) that can be present in drinkingwater without lowering the concentration of F−(aq) below theideal level is closest to(A) 0.25 M(B) 0.025 M(C) 1.6 × 10−6 M(D)Unauthorized copying or reuse ofany part of this page is illegal. GO ON TO THE NEXT PAGE.-24-1.6 × 10−15 M
Question
CaF 2(s) ← Ca2+(aq) + 2 F−(aq) K = 4.0 × 10−11sp49. The concentration of F (aq)− in drinking water that isconsidered to be ideal for promoting dental health is4.0 × 10−5 M. Based on the information above, the maximumconcentration of Ca 2+(aq) that can be present in drinkingwater without lowering the concentration of F−(aq) below theideal level is closest to(A) 0.25 M(B) 0.025 M(C) 1.6 × 10−6 M(D)Unauthorized copying or reuse ofany part of this page is illegal. GO ON TO THE NEXT PAGE.-24-1.6 × 10−15 M
Solution
The solubility product constant (Ksp) expression for the reaction is:
Ksp = [Ca2+][F−]^2
We know that Ksp = 4.0 × 10−11 and [F−] = 4.0 × 10−5 M. We can substitute these values into the Ksp expression to solve for [Ca2+].
4.0 × 10−11 = [Ca2+](4.0 × 10−5)^2
Solving for [Ca2+] gives us:
[Ca2+] = 4.0 × 10−11 / (4.0 × 10−5)^2
[Ca2+] = 4.0 × 10−11 / 1.6 × 10−9
[Ca2+] = 2.5 × 10−2 M
So, the maximum concentration of Ca2+ that can be present in drinking water without lowering the concentration of F− below the ideal level is closest to 0.025 M (Option B).
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