Question

How many moles of NH₄Cl must be added to 100 mL of 0.1 M NH₄OH solution to prevent precipitation of Mn(OH)₂ when this solution is added to 100 mL of a 0.002 M solution of MnCl₂? Assume no change involume on addition of NH₄Cl.

Answer 1

Given

Volume of 0.1 M NH₄OH = 100 mL

Volume of a 0.002 M solution of MnCl₂ =100 mL

Strategy:

We calculate concentration of Mn²⁺ and from the ksp value we also calculate concentration of OH- . Once we know the [OH-] needed to precipitate then that much amount is added to it.

Ksp of Mn(OH)2 is 2.0 E-13

Mol MnCl2 = 0.100 L x 0.002 M = 0.0002

Total volume when NH4OH is added = 0.200 L

[Mn²⁺] = 0.002/0.2 = 0.001 M

Ksp expression

            Mn(OH)2 (s) --- > Mn²⁺ (aq)   + 2OH- (aq)

I                                          0.001                    0

C         -x                                 +x                    +2x

C                                    0.001+x                    2x

Ksp = [Mn²⁺ (aq)] [OH-]²

2.0E-13 = (0.001+x) (2x)²

Since the value of ksp is very small we can neglect x in the denominator.

2.0E-13 = (0.001) (2x)²

x = 7.07 E-6

[OH-] = 2x = 2*7.07 E-6 = 1.41 E-5

Calculation of moles of OH

Mol OH- = 1.41 E-5 M x (1 mol OH-/ 1 mol NH4OH) x 0.200 L

= 2.82 E-6 mol OH-

In order to ppt Mn needs 2.82 E-6 mol

Lets calculate moles of OH- in NH4OH

Mol OH- in NH4OH = 0.1 M x 0.100 L = 0.01 mol

Excess moles of OH- = 0.01 mol OH- - 2.82 E-6 mol

= 0.01 mol OH-

So we need to add more than 0.01 mol of NH4Cl in order to take 0.01 mol OH-

So that will prevent ppt.

Ksp = 4x³

2.0 E-13 = 4x³

x =