Answer:
CaF2 will not precipitate
Explanation:
Given
Volume of Ca(NO3)2 [tex]= 10[/tex] ml
Molar concentration of Ca(NO3)2 [tex]= 0.001[/tex]
Volume of NaF [tex]= 10[/tex] ml
Molar concentration of  NaF  [tex]= 0.0001[/tex]
Ksp for CaF2 [tex]= 3.2 * 10^ {-11}[/tex]
CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2
Moles of calcium ion
[tex]= 10 * 0.001\\= 0.01[/tex]
[tex][Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}[/tex]
Moles of F- ion
[tex]= 10 * 0.0001\\= 0.001[/tex]
[tex][F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}[/tex]
[tex]Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}[/tex]
Q is lesser than Ksp value of CaF2. Hence it will not precipitate
