Question

will a precipitation form when 0.10L of 8.0x10^-3M Pb(N03)2 is added to 0.40L of 5.0x10^-3M Na2So4?

Answer 1

we have

0.10 L of 8.0 x 10^-3 M Pb(N03)2 and 0.40 L of 5.0 x 10^-3 M Na2So4

and after addition concentration of these solutions are

[Pb[NO3)2] = ( 0.10 L x 8.0 x 10^-3 M) / 0.50 L = 1.6 x 10^-3 M

[Na2SO4] =  ( 0.40 L x 5.0 x 10^-3 M) / 0.50 L = 4.0 x 10^-3 M

and the precipitation constan for these two solution is

Q = [Pb2+] [SO4^2-]

= 1.6 x 10^-3 M x  4.0 x 10^-3 M

Q = 6.4 x 10^-6

and

Ksp for the PbSO4 is 1.6 x 10^-8

therefore

Q > Ksp

and

precipitation will occur.