Answer :
Answer:
6.1×10^(-10) mol
Explanation:
CaF2(s) -----------> Ca2+(aq) +2F-(aq)
Ksp = [Ca2+][F-]^2
Given data
[F-] = 0.255 M
∴ The concentration of Ca2+required to start the precipitation
[Ca2+] = Ksp / [F-]2
= (4.0 X 10^-11) /(0.255)^2
= 6.1×10^(-10) M
Here the volume is 1.0 L
∴ The number of molesof Ca(NO3)2 must be added to 1.0 L of a1.0 L of a 0.255 M HF solution to begin precipitation ofCaF2 (s)= 6.1×10^(-10) mol