Chemistry Solubility Math Question 3

Question

A pipette is used to add 49 mL of 0.4 M Pb(NO₃)₂ to a 273.0 mL solution of 0.2 M NaF. Determine the equilibrium concentration of Pb²⁺(aq) in the final solution. K_sp(PbF₂)=3.6E-8

Answer 1

49 ml x 1 L / 1000 ml x 0.4 mol/L = 0.0196 moles Pb²⁺

273 ml x 1 L / 1000 ml x 0.2 mol/L = 0.0546 moles F^-

Final volume = 49 ml + 273 ml = 322 ml = 0.322 L

Final [Pb^2+] = 0.0196 moles/0.322 L = 0.0609 M Pb²⁺

Final [F^-] = 0.0546 moles/0.322 L = 0.1696 M F^-

Q = [Pb²+][F^-]2 = (0.0609)(0.1696)² = 0.00175

Q >> Ksp so a precipitate of PbF₂ will form

Pb²⁺(aq) + 2F^-(aq) <===> PbF₂(s)

0.0609.........0.170.....................0.........Initial

-0.0609.......-0.1218.................+0.1218....Change

0....................0.0482...............0.1218.....Equilibrium

Ksp = [Pb²⁺][F^-]²

3.6x10^-8 = (x)(0.0482]² = 0.00232x

x = 1.56x10^-5 M = [Pb²⁺]