Question

Calculate Qsp for calcium fluoride (Ksp= 3.9x10^-11) when 160.0 mL of a .00565 M solution of Ca(NO3)2 is added to 190.0 mL of a .00650 M solution of KF.

Answer 1

(Ksp= 3.9x10^-11)

[Ca^+2] = 160 x0.00565/1000=9.04 x01^-4 moles in Ca(NO3)2

[F-] = 0.00650 x190/1000 = 1.23 x10^-3 moles in KF

Qsp = [Ca^+2][F-]

     = 9.04 x01^-4 x 1.23 x10^-3

     =11.11 x10^-7

      = 1.11 x10^-6

here Qsp > Ksp

Answer 2

Ca(NO₃)₂ ----------->   Ca²⁺ + 2NO₃^-

moles of Ca(NO₃)₂ = 0.00565 *160/1000 = 0.904 *10^-3

[Ca2+] = 0.904 *10^-3 moles

molarity = 0.904 *10^-3 moles/ total volume in L = 0.904 *10^-3 moles/0.35 = 2.58*10^-3 M

KF ------> K⁺ +   F^-

moles of KF = 0.0065*190/1000 = 1.235 *10^-3

[F-] = 1.235 *10^-3

Molarity = 1.235 *10^-3 /0.35 = 3.53 *10^-3 M

CaF₂ --------> Ca²⁺ +    2F^-

Qsp = [Ca2+][F]²

       = 2.58*10^-3 M * (3.53 *10^-3)² = 32.15 * 10^-9 M