A precipitate of lead(II)chloride forms when 150.0 mg of NaCl is dissolved in 0.250 L of 0.12 M lead(II)nitrate. True or False? Ksp of PbCl2 is 1.7 x 10-5.

Question

A precipitate of lead(II)chloride forms when 150.0 mg of NaCl is dissolved in 0.250 L of 0.12 M lead(II)nitrate. True or False? K_sp of PbCl₂ is 1.7 x 10^-5.

The answer is: false

please explain

Answer 1

To determine if a precipitate forms, we will calculate the reaction quotient (Q) and compare it to the Ksp.

Calculation of Q:

2NaCl(aq) + Pb(NO₃)₂(aq) ===> 2NaNO₃(aq) + PbCl₂(s)

moles NaCl = 150.0 mg x 1 g/1000 mg x 1 mole/58.5 g = 0.002564 moles Cl^-

[Cl-] = 0.002564 mol / 0.25 L = 0.0103 M Cl-

[Pb²⁺] = 0.12 M Pb²⁺

Q = [Pb²⁺][Cl^-]²

Q = (0.12)(0.0103)² = 1.27x10^-5

Compare this to Ksp value of 1.7x10^-5. Q is greater than Ksp, so a precipitate will form. So the statement is TRUE.