Answer to Question #286482 in Chemistry for khryst

Question

1.     What is the solubility product expression for silver carbonate?

a.      K_sp = [Ag⁺]² [CO₃²⁻]

b.     K_sp = [Ag⁺] [CO₃²⁻]²

c.      K_sp = 2 [Ag⁺] [CO₃²⁻]

d.     K_sp = ½ [Ag⁺] [CO₃²⁻]

 

      

3.     What is the molar solubility of the solution? (2 points)

a.      3.23 × 10⁻⁴

b.     2.24 × 10⁻³

c.      1.30 × 10⁻³

d.     1.28 × 10⁻⁴

 

    

5.     Calculate the mass of silver carbonate that will dissolve in 150.0 mL of water. (2 points)

a.      0.0134 g

b.     0.0927 g

c.      5.29 × 10⁻³ g

d.     0.0538 g

 

Answer 1

Solution:

(1) What is the solubility product expression for silver carbonate?

silver carbonate = Ag₂CO₃

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq)

Thus, the K_sp expression for Ag₂CO₃(s) is:

K_sp(Ag₂CO₃) = [Ag⁺]² [CO₃²⁻]

Answer (1):

a) K_sp = [Ag⁺]² [CO₃²⁻]

(3) What is the molar solubility of the solution?

K_sp(Ag₂CO₃) = 8.10×10⁻¹² (the table value)

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq)

___S__________2S_________S_______

K_sp(Ag₂CO₃) = [Ag⁺]² [CO₃²⁻] = [2S]² [S] = 4S³ = 8.10×10⁻¹²

4S³ = 8.10×10⁻¹²

S³ = 2.025×10⁻¹²

S = 0.0001265 = 1.27×10⁻⁴

The molar solubility of the solution is 1.27×10⁻⁴ M

Answer (3):

d) 1.28×10⁻⁴

(5) Calculate the mass of silver carbonate that will dissolve in 150.0 mL of water.

S(Ag₂CO₃) = 1.27×10⁻⁴ M = 1.27×10⁻⁴ mol/L

Volume of water = 150.0 mL = 0.15 L

The molar mass of Ag₂CO₃ is 275.7453 g/mol

Hence,

m(Ag₂CO₃) = S(Ag₂CO₃) × M(Ag₂CO₃) × V

m(Ag₂CO₃) = (1.27×10⁻⁴ mol L⁻¹) × (275.7453 g mol⁻¹) × (0.15 L) = 0.005253 g = 5.25×10⁻³ g

m(Ag₂CO₃) = 5.25×10⁻³ g

Answer (5):

c) 5.29×10⁻³ g