I following are the answers, but I need the work 1) [Ag+] = 1.34x10^-5M, [Cl] =

Question

I following are the answers, but I need the work

1) [Ag+] = 1.34x10^-5M, [Cl] = 1.34x10^-5 M

2) [Pb] = 1.62x10^-2M, [Cl] = 3.24x10^-2

3) [Ba] = 2.5x10^-5M, [F] = 0.2M

4) [Ca] = 2.0x10^-3M, [SO4] = 0.012M

5) Lead Iodide

Fundamental of Analytical Chemistry CHE 311 PROBLEM SET V 1. Silver chloride is a slightly soluble material in water. Calculate the concentration of silver ions and chloride ions at equilibrium in pure water. 2. LeadCI) chloride is a slightly soluble material in water. Calculate the concentration of lead ions and chloride ions at equilibrium in pure water. 3. Barium fluoride is a slightly soluble material in water. Calculate the concentration of barium ions and fluoride ions at equilibrium in a 0.2 M solution of sodium fluoride (neglect ionic strength effects, Ksp 1.0x 10A6) 4. Calcium sulfate is a sparingly soluble material in water. Calculate the concentration of calcium ions and sulfate ions at equilibrium in a 0.01 Mpotassium sulfate solution (neglect ionic strength solution). 5. Which compound is more soluble in pure water, lead(II) iodide or strontium sulfate.

Explanation / Answer

1)

Ksp of AgCl = 1.77 x 10^-10

AgCl ---------------------> Ag+ + Cl-

S S

Ksp = [Ag+][Cl-]

Ksp = S^2

1.77 x 10^-10 = S^2

S = 1.34 x 110^-5 M

S = [Ag+] = [Cl-] = 1.34 x 110^-5 M

2)

Ksp of PbCl2 = 1.70 x 10^-5

PbCl2 ----------------------> Pb+2 + 2Cl-

S 2S

Ksp = [Pb+2][Cl-]^2

Ksp = (S)(2S)^2

Ksp = 4S^3

1.70 x 10^-5 = 4 S^3

S = 1.62 x 10^-2 M

[Pb+2] = S =  1.62 x 10^-2 M

[Cl-] = 2S = 2 x 1.62 x 10^-2 = 3.24 x 10^-2 M

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