Based on the information above, the maximum concentration of Ca2+(aq) that can be present in drinking water without lowering the concentration of F−(aq) below the ideal level is closest to

Question

CaF₂(s)⇄Ca²⁺(aq)+2F⁻(aq)          Ksp=4.0×10⁻¹⁰

The concentration of F⁻(aq) in drinking water that is considered to be ideal for promoting dental health is 4.0×10⁻⁵M. Based on the information above, the maximum concentration of Ca²⁺(aq) that can be present in drinking water without lowering the concentration of F⁻(aq) below the ideal level is closest to

1. 0.025 M
2. 1.6 x 10^-15 M

1. 0.25 M

Answer 1

CaF2(s) ==> Ca²⁺(aq) + 2F^-(aq)

Ksp = [Ca²⁺][F-]²

4.0x10^-10 = [Ca²⁺][4.0x10^-5]²

[Ca²⁺] = 4.0x10^-10 / 1.6x10^-9

[Ca²⁺] = 0.25 M