In: Chemistry
Will a precipitate of magnesium fluoride form when 200.0 ml of 1.9 x 10-3 M MgCl2 are added to 300.0 ml of 1.4 x 10-2 M NaF (Ksp (MgF2 = 6.9 x 10-9 )
Please explain, thanks!
Answer – We are given, 200.0 ml of 1.9 x 10-3 M MgCl2 and 300.0 ml of 1.4 x 10-2 M NaF, , Ksp MgF2 = 6.9 x 10-9
We know when we added the MgCl2 and NaF there is formed MgF2, so there is formed precipitation or not we need to calculate the Qsp
We know expression for the Qsp
Qsp = [Mg2+][F-]2
Now we need to calculate the[Mg2+] and [F-] after mixing the solution.
Total volume = 200 +300 = 500 mL
Moles of MgCl2 = moles of Mg2+ = 1.9 x 10-3 M * 0.200 L
= 0.00038 moles
Moles of NaF = moles of F- = 1.4 x 10-2 M * 0.300 L
= 0.0042 moles
So, [Mg2+] = 0.00038 moles / 0.500 L
= 0.00076 M
[F-] = 0.0042 moles / 0.500 L
= 0.0084 M
So, Qsp = [Mg2+][F-]2
= (0.00076) (0.0084)2
= 5.36*10-8
So, Qsp > Ksp and we know when Qsp > Ksp then there is precipitation occurred, so magnesium fluoride will a precipitate