Question

(13) will a precipitate of Mg F2 form when 300 mL of 1.10*10^-3M MgCl2 solution is added to 500, mL of 1.20 *10^-3M NaF? The Ksp of the precipitate ( if one forms) is 6.9*10^-9.

Answer 1

Lets find the concentration after mixing for MgCl2

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

M(MgCl2) after mixing = M(MgCl2)*V(MgCl2)/(total volume)

M(MgCl2) after mixing = 0.0011 M*300.0 mL/(300.0+500.0)mL

M(MgCl2) after mixing = 4.125*10^-4 M

Lets find the concentration after mixing for NaF

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

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

M(NaF) after mixing = 0.0012 M*500.0 mL/(500.0+300.0)mL

M(NaF) after mixing = 7.5*10^-4 M

So, we have now

[Mg2+] = 4.125*10^-4 M

[F-] = 7.5*10^-4 M

At equilibrium:

MgF2 <----> Mg2+ + 2 F-

Qsp = [Mg2+][F-]^2

Qsp = (4.125*10^-4)*(7.5*10^-4)^2

Qsp = 2.32*10^-10

we have,

Ksp = 6.9*10^-9

Since Qsp is less than ksp, precipitate will not form