10.0 mL of 0.0040 mol/L Pb(NO3)2 (aq) is added to 15.0 mL of 0.25 mol/L KCl(aq). If Ksp = 1.2x10^−5 for PbCl2 will a precipitate form?

Question

10.0 mL of 0.0040 mol/L Pb(NO3)2 (aq) is added to 15.0 mL of 0.25 mol/L KCl(aq). If Ksp = 1.2x10^−5 for PbCl2 will a precipitate form?

a)A precipitate will not form

b)The system is at equilibrium

c)There is not enough information given

d)A precipitate will form

Answer 1

moles Pb present:

0.01 L x 0004 mol/L = 4x10-5 moles Pb²⁺

moles Cl- present:

0.015 L x 0.25 mol/L = 3.75x10^-3 moles Cl-

[Pb²⁺] = 4x10^-5 mol/0.025 L = 1.60x10^-3 M

[Cl-] = 3.75x10^-3 mol/0.025 L = 0.15 M

Q = [Pb²⁺][Cl-]² = (4x10^-5)(0.15)²

Q = 1.78x10^-3

Since Q > Ksp a precipitate will form

Answer 2

The important part of the chemical reaction is: Pb + 2Cl -> PbCl2

To solve this problem, we must determine if the concentrations of both Pb and Cl2 are greater than the Ksp.

Multiplying volumes and molarities, we get:

0.00004 mol Pb

0.00375 mol Cl

Ksp = [Cl2]²[Pb]

Ksp is an equilibrium constant and if the right sight of the equation is greater than the left, a precipitate will form. Vice versa.

Ksp = 0.000012

the concentrations of cl and pb need to be readjusted for their new volumes

moles of cl2/.025L = new cl2 molarity

moles of Pb/.025 L = new Pb molarity

Cl M = .15 M

Pb M = .016 M

[cl]^2[Pb} = 0.00036 = Q

Ksp < Q : Therefore, precipitate forms