CHM 2046C Module 12 Name: ______ Sample _____

Part L. Common Ion Effect on Solubility 10 points

What is the solubility, in milligrams per milliliter, of BaCO₃, K_sp =1.2 x 10 ^-10, in

(a) pure water and

BaCO₃ (s) + HOH (l) ß à Ba²⁺ (aq) + CO₃^2- (aq)

K_sp = [Ba²⁺][CO₃^2-] K_sp = 5.6 x 10 ^-12

Let x = solubility of Barium Carbonate in water, then

(x) (x) = 5.6 ∙10 ^-12

x² = 5.6 ∙10 ^-12

x = [Ba²⁺] = [CO₃^2-] = 2.4 ∙10 ^-6 M

Now find the solubility in mg/mL:

Or 0.00047 mg/mL

(b) in water containing 10.0 mg/ml of Na₂CO₃? Also what is the pH?

The carbonate ion becomes a common ion in the equilibrium:

BaCO₃ (s) + HOH (l) ß à Ba²⁺ (aq) + CO₃^2- (aq)

Now change the 10.0 mg/ml of Na₂CO₃ to its Molarity:

1^st find the molar mass of Na₂CO₃

2^nd take the 10.0 mg/ml of Na₂CO₃ and transform it to its Molarity

3rd recalculate the Molar Solubility of the Barium Carbonate with the Common Ion:

K_sp = [Ba²⁺][CO₃^2-] K_sp = 5.6 x 10 ^-12

(x) (x+0.0943) = 5.6 ∙10^-12

Assume the x in (x+0.0943) is so small it can be neglected!

Let x equal the Molar Solubility of the BaCO₃:

0.0943x = 5.6 ∙10^-12

x = 59.4 ∙10 ^-12 = 5.94 ∙10^-11 M

Assumption is Valid because the value x = 5.94 ∙10^-11 in (x+0.0943) is so small it can be neglected!

The presence of the common ion suppresses the concentration of the Barium ion

and the Molar Solubility of the Barium Carbonate.

(c) What is the pH of the Solution?

The Sodium and the Bariun ions do not react with water (both strong bases). They are spectators.

The Carbonate ion does hydrolyze with water as follows:

CO₃^2- + HOH ↔ HCO₃^1- + OH^1- K_a2HCO3 = 4.8 x 10^-11

K_w [HCO₃^1-] [OH ^1-]

K_h = ---- = ----------------------

K_a [CO₃^2-]

1.0 x 10^-14 [HCO₃^1-] [OH ^1-]

2.1 x 10^-4 = -------------- = ----------------------

4.8 x 10^-11 [CO₃^2-]

	[HCO3^1-]	[OH^-1]	[CO3^2-]
Initial	0	0	0.094
Change	x	x	- x
Equilibrium	x	x	0.094-x

[HCO₃^1-] [OH ^1-] x ∙x

2.1 x 10^-4 = --------------------- = ---------

[CO₃^2-] 0.094-x

Assume x in 0.0943 – x is small so you can neglect it.

Therefore

(0.0.094)( 2.1 x 10^-4) = x²

2.0 x 10^-5 = 20. x 10^-6 = x²

4.4 x 10^-3 = x

Assumption incorrect: 0.094 – 0.0044 is small, but assumption is almost valid

Therefore

2.1 x 10^{-4 (}0.094-x) = x²

2.0 x 10^{-6 -} 2.1 x 10^-4 x = x²

x² + 2.1 x 10^-4 x - 2.0 x 10^-6 = 0

using the quadratic formula x = 0.0016 M = [OH ^1-]

pOH = - log(0.0016 ) = 2.80

pH = 14.00-pOH

pH = 14.00-2.80 = 11.20

Read Section 15.12 in McMurry pages 622-624. Syudy Worked example 15.11 on page 623. Try problem 15.25 on Page 623. Also work Problems 15.100-15.103 at the end of the chapter on page 642.

Additional Worked Example: