CHEM 2 Solubility

Question

Calculate the Qsp or Ksp, as indicated, and determine whether a precipitate will form when each of the following mixtures is prepared. (You may assume temperature is held constant, volumes are additive, and that the initial solutions, prior to mixing, are unsaturated.)

(a) 25.09 mL 1.30 ✕ 10−4 M CaCl2 is mixed with 25.25 mL 3.08 ✕ 10−3 M NaF. (Ksp for CaF2 = 3.5 ✕ 10−11)

Qsp

WHich is correct?

A precipitate will form.

A precipitate will not form.

Answer 1

Let's look at the balanced chemical equation for the reaction taking place when CaCl₂ is mixed with NaF.

CaCl₂(aq) + 2NaF(aq) <==> CaF₂(s) + 2NaCl(aq)

To determine if a precipitate of CaF₂ will form, we will calculate the Qsp and compare it to the Ksp. If the Qsp is less than the Ksp, no precipitate will form, and if Qsp is greater than Ksp, a precipitate will form.

To calculate Qsp, we look at the dissolution of CaF₂ as follows:

CaF₂(s) <==> Ca²⁺(aq) + 2F^-(aq)

Qsp = [Ca²⁺][F-]²

So now, we will calculate these concentrations so we can plug them into the equation for Qsp....

moles Ca²⁺ = 25.09 ml x 1 L / 1000 ml x 1.30x10^-4 mol / L = 3.26x10^-6 moles Ca²⁺

moles F^- = 25.25 ml x 1 L / 1000 ml x 3.08x10^-3 mol / L = 7.78x10^-5 moles F^-

Final volume = 25.09 mls + 25.25 mls = 50.34 mls = 0.05034 L

[Ca²⁺] = 3.26x10^-6 moles Ca²⁺ / 0.05034 L = 6.48x10^-5 M

[F^-] = 7.78x10^-5 moles F^- / 0.05034 L = 1.55x10^-3 M

Plug these into the Qsp expresssion ...

Qsp = (6.48x10^-5)(1.55x10^-3)² = 1.56x10^-10

Compare this value to Ksp of 3.5x10^-11

Qsp > Ksp therefore a precipitate of CaF₂ WILL FORM