Question

3. Insoluble metal hydroxides can be used as a "buffer" to resist changes in pH. For example, consider the solution that would sul if 0.200 moles of MgCl2 are combined wih0.200 moles of NaOH in a total volume of 1.00 liter a) Determine the concentrations of all ionic species in this solution b) Calculate the pH of this solution c) You could add 0.040 moles of NaOH to the solution in part (a), causing additional Mg(OH)2 to precipitate. Calculate the new pH that would result from this addition d) Or, you could add 0.040 moles of HCI to the solution in part (a), which would cause some of the Mg(OH)2 to dissolve. Calculate the new pH that would result from this addition. You should notice that the pH will only change slightly when a reasonable quantity of strong base or strong acid was added to this solution in parts (c) and (d Kw = 1.0 × 10-14 Ka (HF) = 6.6 x 10- 4 Ka (HCN) 4.9x10-10 KatH3Cit) = 7.4 × 10-4 2(HCit)-1.7 x 10-5 K(HCit 4.0 x 10-7 Kip (Cd(OH)2) = 2.5 10-14 Ksp (ScF3) 4.2 x 10-8 Ksp (Zn(OH)2) = 1.2 × 10-17 Ksp (Mg(OH2)-1.8x 10-11 Ksp (Fe(OH)3) 4,0 x 10-38 Ksp (AgCl) = 1.6 x 10-10 Ksp (Ag2G04) = 2.6 x 10-12 Ky(Zn(OH)42-) = 4.6× 1017 Ky (Fe(Cit)) = 6.3x 1011 Ky (Ag(CN)2" ) = 1.0 × 1021

Answer 1

a) Mg2+ + 2OH- ====> Mg(OH)2(s)

We have 0.2 moles of Mg2+ and 0.2 moles of OH- which react to form Mg(OH)2, so finally we have 0.1 moles of Mg2+ left.

Now Mg(OH)2 has Ksp 1.8 × 10-11

Mg(OH)2 ====> Mg2+ + 2OH-

From Ksp, [OH-] = 1.35 × 10-5 M while [Mg2+] = 0.1 M,

[Cl-] = 0.4M, [Na+] = 0.2M

B) pH = 14 + Log[OH-] = 9.127

C) Now we have 0.100 moles of Mg2+. Addition of 0.04 moles of OH- will react with 0.02 moles of Mg2+ and left 0.080 moles of Mg2+. Now

Mg(OH)2 ====> Mg2+ + 2OH-

[OH-] = 0.000015 M, [Mg2+] = 0.08M, [ Na+] = 0.14 M, [Cl-]= 0.4M,

pH = 9.176

D) Now Mg(OH)2 ====> Mg2+ + 2OH-

Now OH- + H+ ----> H2O

Now 0.04 moles of H+ react with 0.04 moles of OH-

Which creat 0.02 moles of Mg

Now we have 0.12 moles of Mg,

From Ksp [OH-] = 1.22 × 10-5 M

pH = 9.088