Question

9. (a) 100 mg of lead (1) hydroxide is added to 1.0 L of water. Does the solid completely dissolve? What is the pH7 (b) What is the minimum amount of milliers of 1,00 M sodium hydroxide must be added to dissolve the precipitate? (12 pts.) No it does not Pb (OH)₂ 15 = P6tean + 204 cm completely dissolve t2x kap=1.2 X 10-15 H UH 2x 1.2X10-15 = (x) (2x)² = 4 x3 the X= 6.69x10-6 the pH = 14-pot = 14 + log (2 x 669 x100) = 9.13 1,2x10-15 - [Phet] [OH)² 11,2x10-15 LOH JE V 6.69x10-6 = 1.34 X 10 M to (1.34x100 m ) (1.0L)= 1.34x10-5 mol it oH 1.34x10-5 mol of DH = 1.34x10-5 L of NaOH 1.oom = 0.0134 mL of Molly 1.34x10 2= /2 Page 6 of 10

Answer 1

The dissociation of Lead Hydroxide (Pb(OH)2) in water takes place as

Pb(OH)2 ----------> Pb⁺² + 2OH^-

We know the solubility product (Ksp) for Pb(OH)2 = 1.2 * 10^-15

That means [Pb⁺²] * [OH^-] = 1.2 * 10^-15

Now given 100 mg of Pb(OH)2 is added to 1 L of water

That means moles of Pb(OH)2 added = mass/ molar mass

= 100 mg/ 241.21 g/mol

= 0.1 g/ 241.21 g/mol

= 0.000415 moles

Thus initial concentration of Pb(OH)2 added = moles / volume

= 0.000415 moles / 1 L

= 0.000415 M

Thus if we form the ICE table, then

Reaction	Pb(OH)2	Pb+2	2OH-
Initial	0.000415 M	0	0
Change	-x	+x	+2x
Equilibrium	0.000415 -x	x	2x

Thus Ksp = [Pb⁺²] * [OH^-]²

= [x] * [2x]²

= 4x³   

i.e 1.2 * 10^-15 = 4x³

x³   = 1.2 * 10^-15 / 4

= 0.3 * 10^-15

   x = (0.3 * 10^-15 )^1/3  

= 0.669 * 10^-5

= 6.69 * 10^-6

Thus [OH^-] = 2x = 2* (6.69 * 10^-6) = 13.38 ‬* 10^-6

Hence pOH = -log [OH^-] = -log [13.38 ‬* 10^-6] = 6 - log 13.38 = 6 - 1.126 = 4.874‬

And pH = 14 - pOH = 14 - 4.874‬ = 9.126‬

2- If the Initial [Pb(OH)2] =  0.000415 M

Then the reaction is said to be saturated when all the Pb(OH)2 is converted to Pb+2 and OH- completely.

Hence the reaction quotient Q for  Pb(OH)2 ----------> Pb⁺² + 2OH^-

is Q = [Pb⁺²] * [OH^-]²

= [0.000415] * [2*0.000415]²

= [0.000415] * [0.00083‬]²

= 2.85 * 10^-10

Now since Q > Ksp, that means the precipitation will take place until Q becomes equal to Ksp

Since Q > Ksp that means we have OH^- in excess in product. Now to adjust the OH- with same number in reactant, we have to add NaOH on reactant.

The excedd OH- in product = 0.00083 - (13.38 ‬* 10^-6 )

= 0.00081662 moles

Thus NaOH added should be = 0.00081662 moles

Now given concentration of NaOH added = 1.00M

Thus required volume of NaOH = moles/ concentration

= 0.00081662 moles / 1.00M

= 0.00081662 moles / 1.mole/ 1000 ml

= 0.816 ml