Insoluble metal hydroxides can be used as a \"buffer\" to resist changes in pH.
ID: 1031472 • Letter: I
Question
Insoluble metal hydroxides can be used as a "buffer" to resist changes in pH. For example, consider the solution that would result if 0.200 moles of MgCl2 are combined with 0.200 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 HCl 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)!Explanation / Answer
MgCl2+2NaOH--------->Mg(OH)2+2NaCl
Mg(OH)2--------->Mg2++2OH-
(a)
From first reaction NaOH is limiting reactant.
So as per stoichiometry
NaOH=0.2 moles
MgCl2=0.1 moles
Mg(OH)2=0.1 moles
So in 1 lit solution
Mg(OH)2=0.1 moles/lit=0.1 M
So
Mg2+=0.1 M
(b)
Now Ksp of Mg(OH)2 is 1.8 x 10-11
Ksp=[Mg2+][OH-]2=0.1*[OH-]2=1.8 x 10-11
[OH-]=1.34*10-5
pOH=-log[OH-]=4.872
pH=14-4.872=9.128
(c)
Now if 0.04 moles of NaOH added
According reaction
NaOH=0.24 moles
MgCl2=0.12 moles
Mg2+ = 0.12 M
So Ksp=[Mg2+][OH-]2=1.8 x 10-11
[OH-]=1.22*10-5
pOH=4.912
pH=9.088
(d)
if 0.04 moles HCl added
then MgCl2=0.1-0.02=0.08 moles
Mg2+=0.08 M
So,Ksp=[Mg2+][OH-]2=1.8 x 10-11
[OH-]=1.5*10-5
pOH=4.823
pH=9.176