Question

Calculate Q_sp for calcium fluoride (K_sp = 3.9 ×10^–11) when 125.0 mL of a 5.35×10^-3M solution of Ca(NO₃)₂ is added to 235.0 mL of a 7.00×10^-3M solution of KF.

Answer 1

Lets find the concentration after mixing for Ca(NO3)2

Concentration after mixing = mol of component / (total volume)

M(Ca(NO3)2) after mixing = M(Ca(NO3)2)*V(Ca(NO3)2)/(total volume)

M(Ca(NO3)2) after mixing = 0.00535 M*125.0 mL/(125.0+235.0)mL

M(Ca(NO3)2) after mixing = 1.858*10^-3 M

Lets find the concentration after mixing for KF

Concentration after mixing = mol of component / (total volume)

M(KF) after mixing = M(KF)*V(KF)/(total volume)

M(KF) after mixing = 0.007 M*235.0 mL/(235.0+125.0)mL

M(KF) after mixing = 4.569*10^-3 M

So, we have now

[Ca2+] = 1.858*10^-3 M

[F-] = 4.569*10^-3 M

At equilibrium:

CaF2 <----> Ca2+ + 2 F-

Qsp = [Ca2+][F-]^2

Qsp = (1.858*10^-3)*(4.569*10^-3)^2

Qsp = 3.879*10^-8

we have,

Ksp = 3.9*10^-11

Since Qsp is greater than ksp, precipitate will form

Answer: QSp = 3.9*10^-8