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

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8