Problem #1: Calculate the milligrams of silver carbonate in first 100 mL (then 250 mL) of a saturated solution of Ag_{2}CO_{3} (K_{sp} for silver carbonate = 8.4 x 10¯^{12})

Solution:

1) Solve for the molar solubility of Ag_{2}CO_{3}:

Ag_{2}CO_{3} <==> 2Ag^{+} + CO_{3}^{2}¯

K_{sp} = [Ag^{+}]^{2} [CO_{3}^{2}¯]

8.4 x 10¯^{12} = (2x)^{2} (x)

x = 1.28 x 10¯^{4} mol/L

Comment: The value of x is the molar concentration of the carbonate ion. Since there is a 1:1 molar ratio between it and silver carbonate, the value for x is also the molar solubility of silver carbonate.

2) Convert mol/L to gram/L:

1.28 x 10¯^{4} mol/L times 275.748 g/mol = 3.53 x 10¯^{2} g/L

3) Convert from g/L to g/100mL:

3.53 x 10¯^{2} g/L divided by 10 = 3.53 x 10¯^{3} g/100mL

Comment: this is done because there are 10 100mL amounts in 1 L.

4) Convert mol/L to g/250mL:

3.53 x 10¯^{2} g/L divided by 4 = 8.83 x 10¯^{3} g/250mL

Comment: this is done because there are 4 250mL amounts in 1 L.