Calculate the milligrams of silver carbonate in first 100 mL (then 250 mL) of a saturated solution of Ag2CO3 (Ksp for silver carbonate = 8.4 x 10¯12)

Question: 

 

Problem #1: Calculate the milligrams of silver carbonate in first 100 mL (then 250 mL) of a saturated solution of Ag2CO3 (Ksp for silver carbonate = 8.4 x 10¯12)

Solution:

1) Solve for the molar solubility of Ag2CO3:

Ag2CO3 <==> 2Ag+ + CO32¯

Ksp = [Ag+]2 [CO32¯]

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.

 

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