Mastering Solubility Equilibrium Calculations
Expert reviewed • 22 November 2024 • 5 minute read
VIDEO
Introduction
Understanding solubility equilibrium calculations is fundamental in chemistry, particularly when dealing with precipitate formation and ionic solutions. This guide walks through common problem types you'll encounter in HSC Chemistry's Solution Equilibrium section.
Key Concepts
The solubility product constant (K s p K_{sp} K s p
Calculate ion concentrations in saturated solutions
Predict precipitate formation
Determine solubility under various conditions
Example Problems and Solutions
1. Ion Concentrations in Saturated Solutions
Problem: Calculate the concentrations of Ba²⁺ and OH⁻ ions in a saturated Ba(OH)₂ solution at 25°C.
K s p = 5.0 × 1 0 − 3 K_{sp} = 5.0 \times 10^{-3} K s p = 5.0 × 1 0 − 3
Solution:
Write the dissociation equation:
Ba(OH) 2 ⇄ Ba 2 + + 2 OH − \text{Ba(OH)}_2 \rightleftarrows \text{Ba}^{2+} + 2\text{OH}^- Ba(OH) 2 ⇄ Ba 2 + + 2 OH −
Let solubility = s s s
[ Ba 2 + ] = s [\text{Ba}^{2+}] = s [ Ba 2 + ] = s [ OH − ] = 2 s [\text{OH}^-] = 2s [ OH − ] = 2 s
Write K s p K_{sp} K s p K s p = [ Ba 2 + ] [ OH − ] 2 = s ( 2 s ) 2 = 4 s 3 K_{sp} = [\text{Ba}^{2+}][\text{OH}^-]^2 = s(2s)^2 = 4s^3 K s p = [ Ba 2 + ] [ OH − ] 2 = s ( 2 s ) 2 = 4 s 3
Solve for ( s ) (s) ( s ) 5.0 × 1 0 − 3 = 4 s 3 5.0 \times 10^{-3} = 4s^3 5.0 × 1 0 − 3 = 4 s 3 s = 0.108 mol L − 1 s = 0.108\text{ mol L}^{-1} s = 0.108 mol L − 1
Therefore:
[ Ba 2 + ] = 0.108 mol L − 1 [\text{Ba}^{2+}] = 0.108\text{ mol L}^{-1} [ Ba 2 + ] = 0.108 mol L − 1 [ OH − ] = 0.216 mol L − 1 [\text{OH}^-] = 0.216\text{ mol L}^{-1} [ OH − ] = 0.216 mol L − 1 2. Precipitate Formation Prediction
Problem: Will a precipitate form when 250.0 mL of 0.150 mol L⁻¹ NaOH is mixed with 250.0 mL of 0.0800 mol L⁻¹ Ba(NO₃)₂?
Solution:
Calculate diluted concentrations:
[ OH − ] = 0.150 mol L − 1 × 250.0 mL 500.0 mL = 0.0750 mol L − 1 [\text{OH}^-] = 0.150\text{ mol L}^{-1} \times \frac{250.0\text{ mL}}{500.0\text{ mL}} = 0.0750\text{ mol L}^{-1} [ OH − ] = 0.150 mol L − 1 × 500.0 mL 250.0 mL = 0.0750 mol L − 1 [ Ba 2 + ] = 0.0800 mol L − 1 × 250.0 mL 500.0 mL = 0.0400 mol L − 1 [\text{Ba}^{2+}] = 0.0800\text{ mol L}^{-1} \times \frac{250.0\text{ mL}}{500.0\text{ mL}} = 0.0400\text{ mol L}^{-1} [ Ba 2 + ] = 0.0800 mol L − 1 × 500.0 mL 250.0 mL = 0.0400 mol L − 1
Calculate ion product (Q):
Q = [ Ba 2 + ] [ OH − ] 2 = 0.0400 × ( 0.0750 ) 2 = 2.25 × 1 0 − 4 Q = [\text{Ba}^{2+}][\text{OH}^-]^2 = 0.0400 \times (0.0750)^2 = 2.25 \times 10^{-4} Q = [ Ba 2 + ] [ OH − ] 2 = 0.0400 × ( 0.0750 ) 2 = 2.25 × 1 0 − 4
Compare with K s p K_{sp} K s p Q < K s p Q < K_{sp} Q < K s p 2.25 × 1 0 − 4 < 5.0 × 1 0 − 3 2.25 \times 10^{-4} < 5.0 \times 10^{-3} 2.25 × 1 0 − 4 < 5.0 × 1 0 − 3
3. Solubility in Common Ion Solutions
Problem: Calculate BaSO₄ solubility in 200.0 mL of 1.00 mol L⁻¹ Na₂SO₄.
K s p of BaSO 4 = 1.1 × 1 0 − 10 K_{sp} \text{ of BaSO}_4 = 1.1 \times 10^{-10} K s p of BaSO 4 = 1.1 × 1 0 − 10
Solution:
Write equilibrium equation:
BaSO 4 ⇄ Ba 2 + + SO 4 2 − \text{BaSO}_4 \rightleftarrows \text{Ba}^{2+} + \text{SO}_4^{2-} BaSO 4 ⇄ Ba 2 + + SO 4 2 −
Initial [ SO 4 2 − ] = 1.00 mol L − 1 [\text{SO}_4^{2-}] = 1.00\text{ mol L}^{-1} [ SO 4 2 − ] = 1.00 mol L − 1
Let solubility = s s s K s p = [ Ba 2 + ] [ SO 4 2 − ] K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] K s p = [ Ba 2 + ] [ SO 4 2 − ] 1.1 × 1 0 − 10 = s ( 1.00 ) 1.1 \times 10^{-10} = s(1.00) 1.1 × 1 0 − 10 = s ( 1.00 )
Solve for s s s s = 1.1 × 1 0 − 10 mol L − 1 s = 1.1 \times 10^{-10}\text{ mol L}^{-1} s = 1.1 × 1 0 − 10 mol L − 1