Answer to Question #327275 in General Chemistry for Anne

Question

1. When a solution of lead(II) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation.

𝑃𝑏(𝑁𝑂₃ )₂ (𝑎𝑞) + 𝐾₂𝐶𝑟𝑂₄ (𝑎𝑞) → 𝑃𝑏𝐶𝑟𝑂₄ (𝑠) + 2𝐾𝑁𝑂₃

(a)What volume of 0.105 M lead (II) nitrate is required to react with 100 mL of 0.120 M potassium chromate

(b)What mass of 𝑃𝑏𝐶𝑟𝑂₄ solid forms?

2. Suppose a titration is run in which 25.05 mL of NaOH solution of unknown concentration reacts with 25.00 mL of 0.1000 M 𝐻₂𝑆𝑂₄ solution. The chemical equation that summarizes the reaction is as follows:

𝐻₂𝑆𝑂₄ (𝑎𝑞) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 2𝐻₂𝑂(𝑙) + 𝑁𝑎₂𝑆𝑂₄(𝑎𝑞)

What is the molarity of the solution?

3. If a solution of 𝐴𝑔𝑁𝑂₃ is added to an HCl solution, insoluble AgCl will precipitate:

𝐴𝑔𝑁𝑂₃ (𝑎𝑞) + 𝐻𝐶𝑙(𝑎𝑞) → 𝐴𝑔𝐶𝑙(𝑠) + 𝐻𝑁𝑂₃(𝑎𝑞)

(a)What volume of 1.20 M 𝐴𝑔𝑁𝑂₃ solution must be added to 250.0 mL of a 0.100 M HCl solution to precipitate all the chloride ion?

(b)What mass of AgCl should precipitate?

Answer 1

1.) 𝑃𝑏(𝑁𝑂₃ )₂ (𝑎𝑞) + 𝐾₂𝐶𝑟𝑂₄ (𝑎𝑞) → 𝑃𝑏𝐶𝑟𝑂₄ (𝑠) + 2𝐾𝑁𝑂₃;

C(𝑃𝑏(𝑁𝑂₃ )₂) = 0.105 M;

C(𝐾₂𝐶𝑟𝑂₄) = 0.120 M;

V(𝐾₂𝐶𝑟𝑂₄) = 100 mL = 0.100 L;

a) C₁ * V₁ = C₂ * V₂

V(𝑃𝑏(𝑁𝑂₃ )₂) = C(𝐾₂𝐶𝑟𝑂₄) * V(𝐾₂𝐶𝑟𝑂₄)/C(𝑃𝑏(𝑁𝑂₃ )₂) = 0.120 * 0.100/0.105 = 0.114 L = 114 mL;

b) n(𝐾₂𝐶𝑟𝑂₄) = C(𝐾₂𝐶𝑟𝑂₄) * V(𝐾₂𝐶𝑟𝑂₄) = 0.120 * 0.1 = 0.012 mol;

n(𝑃𝑏𝐶𝑟𝑂₄) = n(𝐾₂𝐶𝑟𝑂₄) = n(𝑃𝑏(𝑁𝑂₃ )₂) = 0.012 mol;

M(𝑃𝑏𝐶𝑟𝑂₄) = 323 g/mol;

m(𝑃𝑏𝐶𝑟𝑂₄) = n(𝑃𝑏𝐶𝑟𝑂₄) * M(𝑃𝑏𝐶𝑟𝑂₄) = 0.012 * 323 = 3.88 g.

2.) 𝐻₂𝑆𝑂₄ (𝑎𝑞) + 2𝑁𝑎𝑂𝐻(𝑎𝑞) → 2𝐻₂𝑂(𝑙) + 𝑁𝑎₂𝑆𝑂₄(𝑎𝑞)

V(H₂SO₄) = 25.00 mL = 0.02500 L;

C(H₂SO₄) = 0.1000 M;

V(NaOH) = 25.05 mL = 0.02505 L;

C(H₂SO₄) * V(H₂SO₄) = C(NaOH) * V(NaOH)/2;

C(NaOH) = C(H₂SO₄) * V(H₂SO₄) * 2/V(NaOH) = 0.1000 * 25.00 * 2/25.05 = 0.1996 M.

3.) 𝐴𝑔𝑁𝑂₃ (𝑎𝑞) + 𝐻𝐶𝑙(𝑎𝑞) → 𝐴𝑔𝐶𝑙(𝑠) + 𝐻𝑁𝑂₃(𝑎𝑞)

V(HCl) = 250.0 mL = 0.2500 L;

C(HCl) = 0.100 M;

C(𝐴𝑔𝑁𝑂₃) = 1.200 M

a) C(𝐴𝑔𝑁𝑂₃) * V(𝐴𝑔𝑁𝑂₃) = C(HCl) * V(HCl);

V(𝐴𝑔𝑁𝑂₃) = C(HCl) * V(HCl)/C(𝐴𝑔𝑁𝑂₃) = 0.100 * 0.2500/1.2 = 0.0208 = 20.8 mL.

b) n(HCl) = C(HCl) * V(HCl) = 0.100 * 0.2500 = 0.025 mol;

n(AgCl) = n(HCl) = n(𝐴𝑔𝑁𝑂₃) = 0.025 mol;

M(AgCl) = 143 g/mole

m(AgCl) = n(AgCl) * M(AgCl) = 0.025 * 143 = 3.6 g.