Question

For PbCl₂ (K_sp 2.4 x 10^-4), will a precipitate of PbCl₂ form when 0.10 L of 3.0 x 10^-2 M Pb(NO₃)₂ is added to 400 ml of 9.0 x 10^-2 M NaCl?

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Answer 1

Answer – We are given, 0.10 L of 3.0 x 10^-2 M Pb(NO₃)₂ and 400.0 ml of 9.0 x 10^-2 M NaCl, , K_sp PbCl₂ = 2.4 x 10^-4

We know when we added the Pb(NO₃)₂ and NaCl there is formed PbCl₂, so there is formed precipitation or not we need to calculate the Q_sp

We know expression for the Q_sp

Q_sp = [Pb²⁺][Cl^-]²

Now we need to calculate the[Pb²⁺] and [Cl^-] after mixing the solution.

Moles of Pb(NO₃)₂ = moles of Pb²⁺ = 3.0 x 10^-2 M * 0.100 L

                                                        = 0.003 moles

Moles of NaCl = moles of Cl^- = 9.0 x 10^-2 M * 0.400 L

                                             = 0.036 moles

Total volume = 100 +400 = 500 mL

So, [Pb²⁺] = 0.003 moles / 0.500 L

                  = 0.006 M

[Cl^-] = 0.036 moles / 0.500 L

        = 0.072 M

So, Q_sp = [Pb²⁺][Cl^-]²

            = (0.006) (0.072)²

             = 3.11*10^-5

So, Q_sp < K_sp and we know when Q_sp < K_sp then there is no precipitation occurred, so PbCl₂ will not a form precipitate