Insoluble metal hydroxides can be used as a \"buffer\" to resist changes in pH.

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 total volume of 1.00 liter. Determine the concentrations of all ionic species in this solution. W) 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

Following reaction occcur:

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

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