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

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

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

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

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