Calculate the concentration of Pb2+ in a saturated solution of PbCl2. (Ksp = 1.4

Question

Calculate the concentration of Pb2+ in a saturated solution of PbCl2. (Ksp = 1.4 x 10-8)

Explanation / Answer

example Calculate the concentrations of Pb2+ and Cl - at equilibrium? A solution is prepared by mixing 50.0 mL of 0.25 M Pb(NO3)2 with 50.0 mL of 2.2 M KCl. Calculate the concentrations of Pb2+ and Cl - at equilibrium. Ksp for PbCl2(s) is 1.6 10-5. moles Pb2+ = 0.0500 L x 0.25 M = 0.013 moles Cl- = 0.0500 L x 2.2 M = 0.11 total volume = 0.100 L [Pb2+]= 0.013 / 0.100 = 0.13 M [Cl-] = 0.11/ 0.100 = 1.1 M The precipitation reaction Pb2+ + 2Cl- >> PbCl2 requires a 1 - to - 2 ratio so Cl- is in excess Let us assume that the excess Cl- drives all Pb2+ out of solution. This means that 0.13 M Pb2+ reacts with 2 x 0.13 = 0.26 M of Cl- to produce PbCl2 leaving 1.1 - 0.26 = 0.84 M Cl- in excess. Let us now as much PbCl2 dissolve as needed to estabilish the PbCl2 saturation equilibrium let us say , x mol/L. this will rise the concentration of Cl- to 0.84 + 2x and will make the Pb2+ concentration equal to x. PbCl2 Pb2+ + 2Cl- [Pb2+]= x and [Cl-] = 2x + 0.84 Ksp = 1.6 x 10^-5 = (x)( 2x + 0.84)^2 x = 2.3 x 10^-5 [Pb2+] = 2.3 x 10^-5 M and [Cl-] = (2 x 2.3 x 10^-5) + 0.84= 0.84 M

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.