Solid NaCl is added slowly to a solution containing 0.10 M AgNO_3 and 020 M Pb(N

Question

Solid NaCl is added slowly to a solution containing 0.10 M AgNO_3 and 020 M Pb(NO_3)_2. K_sp for AgCl is 1.8 times 10^-10. K_sp for PbCl_2 is 1.6 times 10^-5. a. Write a net ionic equation and corresponding K_sp expression for the dissolution of the following solids: i. silver chloride ii. lead (II) chloride b. Calculate the [Cl^-] required to form each precipitate. c. Which precipitate forms first? Explain your answer. d. What is the concentration of the first metal ion to precipitate when the second one just begins to precipitate? e. If 100 mL of 0.05 M NaCl is added to 200 mL of solution containing 0.005 M AgNO_3 and 0.10 M Pb(NO_3)_2, does a precipitate form? If so, which one(s) form? Explain. A total of 0.450 mol of hydrazoic acid, HN_3 (K_a = 1.9 times 10^-5), is added to enough water to make 1.55 L of solution. a. Write a chemical equation for the reaction of hydrazoic acid with water and write the corresponding equilibrium expression. b. Calculate the pH of the solution. c. Calculate the number of moles of added sodium azide or hydrozoic acid (indicate which) required to make the pH = 5.00. Assume no volume change. d. Calculate the pH of the solution made by adding 0.10 mol of sodium azide to 0.20 mol of hydrazoic acid to make the solution volume 1.00 L. Compare the buffer capacity of this solution to that of the solution prepared in Part c. e. What is the ph of a 0.350 M solution of NaN_3?

Explanation / Answer

ai) Ag+(aq) + Cl-(aq) <--->AgCl(s)

Ksp = [Ag+]e [Cl-]e

aii) Pb+2(aq) + 2Cl-(aq) <--->PbCl2(s)

Ksp = [Pb+]e [Cl-]e2

bi) If s is the concentration of AgCl at equilibrium, then [Ag+]and[Cl-] at equilibrium will be s

Ksp = [Ag+]e [Cl-]e

=s•s =s2

s = Ksp = 1.8 X 10-10

= 1.34 X 10-5 mol L-1

Note, the unit of solubility is mol L-1 or M. However, Ksp is unitless

The [Cl-] required to form AgCl precipitate is 1.34 X 10-5 mol L-1

bii) If s is the concentration of PbCl2 at equilibrium, then [Pb+] at equilibrium will be s and [Cl-] at equilibrium will be 2s

Ksp = [Pb+]e [Cl-]e2

=s•(2s)2 = 4s3

s = 3(Ksp/4) = 3(1.6 X 10-5/4)

= 3(4 X 10-6)

= 1.59 X 10-2 mol L-1

2s = 2 X 1.59 X 10-2 mol L-1 = 3.17 X 10-2 mol L-1

The [Cl-] required to form PbCl2 precipitate is 3.17 X 10-2 mol L-1

c) Referring to Ksp values of AgCl and PbCl2, we know that PbCl2 is more soluble in aqueous medium than AgCl. Also the [Cl-] required to form AgCl precipitate is lower than the [Cl-] required to form PbCl2 precipitate. Hence, AgCl will precipitate first.

d) When PbCl2 begins to precipitate, the [Cl-] will is 3.17 X 10-2 mol L-1

We know Ksp = [Ag+]e [Cl-]e

1.8 X 10-10 = [Ag+]X 3.17 X 10-2

[Ag+]= 1.8 X 10-10 / 3.17 X 10-2

= 0.568 X 10-8

= 5.68 X 10-9 mol L-1

The [Ag+]when PbCl2 begins to precipitate will be 5.68 X 10-9 mol L-1

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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