Question

The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011 M) can be separated by selective precipitation with KOH. What is the concentration of [Mg2+] when Ca(OH)2 starts to precipitate? Ksp Mg(OH)2 = 2.1 x 10-13 and the Ksp Ca(OH)2 = 4.7 x 10-6 2.5 x 10-10 M

Answer 1

the concentration of [Mg²+] is 4.9 × 10⁻¹⁰M

Step-by-step explanation:

Given that,

[Mg²+] = 0.059 M

[Ca²+] = 0.011 M

Ksp Mg(OH)₂ = 2.1 * 10⁻¹³

Ksp Ca(OH)₂ = 4.7 * 10⁻⁶

For precipitation of Ca(OH)₂

[Ca²+] [OH⁻]² = Ksp [Ca(OH)₂]

(0.011) [OH⁻]² = 4.7 × 10⁻⁶

[OH⁻] = 2.067 × 10⁻²M

This is the concentration of [OH⁻]

For [Mg²+] concentration

[Mg²+] [OH⁻]²= Ksp [Mg(OH)₂]

[Mg²+] [2.067 × 10⁻²]² = 2.1 × 10⁻¹³

[Mg²+] = (2.1 × 10⁻¹³) / (4.27 × 10⁻⁴)

[Mg²+] = 4.9 × 10⁻¹⁰M

Therefore, the concentration of [Mg²+] is 4.9 × 10⁻¹⁰M

Answer 2

Ca(OH)2 will precipitate a concentration of [Mg^2+] = 4.9* 10^-10 M

Step-by-step explanation:

Step 1: Data given

([Mg2+] = 0.059 M

[Ca2+] = 0.011 M

Ksp Mg(OH)2 = 2.1 * 10^-13

Ksp Ca(OH)2 = 4.7 * 10^-6 2.5 * 10^-10 M

Step 2: Calculate the concentration of Mg^2+

Q = [Ca^2+][OH-]²

Q = 0.011 * [OH-]²

When Q = Ksp

0.011 *[OH-]² = Ksp = 4.7 * 10^-6

[OH-]² = 4.7 * 10^-6 / 0.011

[OH-]² = 0.000427

[OH-] = 0.0207

Q = [Mg^2+][OH-]²

Q = [Mg^2+] ( 0.0207)²

When Q = Ksp

[Mg^2+]( 0.0207)² = Ksp = 2.1*10^-13

[Mg^2+] = 2.1*10^-13 / ( 0.0207)²

[Mg^2+] = 4.9* 10^-10 M

Ca(OH)2 will precipitate a concentration of [Mg^2+] = 4.9* 10^-10 M