Question

How many grams of Milk of Magnesia, Mg(OH)₂ (s) (58.3 g/mol), would be soluble in 200 mL of water? Ksp = 7.1 × 10⁻¹². Include the ionic reaction and the expression for Ksp in your answer. (Kw = 1 × 10⁻¹⁴ = [H₃O⁺][OH⁻])

a) 0.0032 g
b) 0.0128 g
c) 0.0256 g
d) 0.0512 g

Answer 1

Through the calculation using the Ksp of Mg(OH)2, we find that none of the options given precisely matches the calculated solubility in grams. However, the closest incorrect option provided is 0.0256 g. Option C

Step-by-step explanation:

To determine the amount of Milk of Magnesia, Mg(OH)2 that would be soluble in 200 mL of water, we will use the solubility product constant (Ksp). The dissolution of Mg(OH)2 in water can be represented by the following equilibrium reaction:

Mg(OH)2 (s) → Mg2+ (aq) + 2 OH− (aq)

The expression for Ksp is:

Ksp = [Mg2+][OH−]2

Given the Ksp of 7.1 x 10−12, we can set up the equation:

7.1 x 10−12 = s * (2s)2 = 4s3

where 's' is the solubility of Mg(OH)2 in mol/L. Solving for s gives:

s = (7.1 x 10−12 / 4)1/3 = 1.35 x 10−4 mol/L

Since 1 liter is 1000 mL, for 200 mL the amount of Mg(OH)2 that would dissolve is:

1.35 x 10−4 mol/L * 0.200 L = 2.7 x 10−6 mol

The molar mass of Mg(OH)2 is 58.3 g/mol, so the mass in grams is:

2.7 x 10−6 mol * 58.3 g/mol = 0.15741 x 10−3 g

Truncating to three significant figures:

0.157 g

Therefore, looking at the options given by the student, none of the options correctly match the calculated value. However, closest to the calculated value of 0.157 g is option c) 0.0256 g, although this is not an exact match. Option C