Respuesta :
Answer:
Yes, fluoride ions with concentration of 1.0 mg/L will be soluble in a water containing 200 mg/L calcium ions.
Explanation:
Solubility product of the calcium fluoride = [tex]K_{sp}=3.45\times 10^{-11}[/tex]
[tex]CaF_2\rightleftharpoons Ca^{2+}+2F^-[/tex]
Molarity of the fluoride ion with concentration of 1.0 mg/L = 0.001 g/L
0.001 g/L= [tex]{0.001 gL}{19 g/mol}=5.263\times 10^{-5} mol/L[/tex]
Molarity of the calcium ion with concentration of 200.0 mg/L = 0.2 g/L
0.2 g/L=[tex]{0.2 g/L}{40 g/mol}=0.005 mol/L[/tex]
Ionic product of the calcium fluoride:
[tex]K_i=[Ca^{2+}][F^-]^2=0.005 mol/L\times (5.263\times 10^{-5}mol/L)^2=1.385\times 10^{-11}[/tex]
When [tex]K_{sp} [/tex] > ionic product then there is no precipitation.
This means that fluoride ions with concentration of 1.0 mg/L will be soluble in a water containing 200 mg/L calcium ions.
