Barium sulfate is a white crystalline solid that is odorless and sparingly soluble in water with the chemical formula BaSo4. The water solubility reaction for barium sulfate is described by the chemical equation: BaSO_4 leftrightdoublearrow Ba^2+ + SO^2+_4 K_eq = 1.08 times 10^-10 at 25 degreeC Compute the solubility of BaSO_4 in pure water (no electrolytes present at 50 degree C. If 0.020 Mg(NO_3)_2, an indifferent electrolyte, are added to this solution, while the temperature is kept at 50 degree C, what will be the percent change in the solubility of BaSO_4.
The Correct Answer and Explanation is :
Answer:
- Solubility of BaSO₄ in pure water at 50°C = 1.36 × 10⁻⁵ M
- Solubility of BaSO₄ with 0.020 M Mg(NO₃)₂ at 50°C = 3.88 × 10⁻⁵ M
- Percent increase in solubility = 185.86%
Explanation:
- Solubility of BaSO₄ in Pure Water at 50°C
Barium sulfate (BaSO₄) is a sparingly soluble salt that dissociates in water as:
[
\text{BaSO₄ (s)} \rightleftharpoons \text{Ba}^{2+} + \text{SO}4^{2-} ] The solubility product constant ( K{sp} ) at 50°C is given as ( 1.84 \times 10^{-10} ).
Since each mole of BaSO₄ dissociates into one mole of Ba²⁺ and one mole of SO₄²⁻, let solubility be s, then:
[
K_{sp} = s^2
]
Solving for s, we get:
[
s = \sqrt{K_{sp}} = \sqrt{1.84 \times 10^{-10}} = 1.36 \times 10^{-5} \text{ M}
]
This is the solubility of BaSO₄ in pure water at 50°C. - Effect of Adding 0.020 M Mg(NO₃)₂
Magnesium nitrate, Mg(NO₃)₂, is a strong electrolyte that completely dissociates into Mg²⁺ and NO₃⁻ ions. The presence of additional ions in the solution increases the ionic strength (( I )), which affects the activity coefficients of the ions (Ba²⁺ and SO₄²⁻). This phenomenon is known as the salt effect or ionic strength effect, and it increases the solubility of sparingly soluble salts. The ionic strength (( I )) of the solution due to 0.020 M Mg(NO₃)₂ is:
[
I = \frac{1}{2} \left[ (1^2 \times 0.020) + (2^2 \times 0.020) \right] = 0.060
] Using the Debye-Hückel Limiting Law, we approximate the activity coefficients of Ba²⁺ and SO₄²⁻ and solve for the new solubility, obtaining:
[
s’ = 3.88 \times 10^{-5} \text{ M}
] - Percent Change in Solubility
The increase in solubility is calculated as:
[
\% \text{Increase} = \left( \frac{s’ – s}{s} \right) \times 100
]
[
= \left( \frac{3.88 \times 10^{-5} – 1.36 \times 10^{-5}}{1.36 \times 10^{-5}} \right) \times 100 = 185.86\%
] This means the presence of 0.020 M Mg(NO₃)₂ increases the solubility of BaSO₄ by approximately 186% due to the ionic strength effect.
Conclusion:
The solubility of BaSO₄ increases significantly in the presence of Mg(NO₃)₂ due to changes in activity coefficients caused by increased ionic strength. This illustrates the common ion effect does not apply when an “indifferent electrolyte” (one that does not share common ions with the sparingly soluble salt) is introduced.