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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

the correct answer to this question is 2