Please help me with this topic as I am very confused and I\'m even working with

Question

Please help me with this topic as I am very confused and I'm even working with someone for help. Please show all work and write neatly if possible. Thank you so very much.

Note: I have posted equilibrium constants (below the problem) needed to answer the problem. Thank you.

3. 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. Mg(OH)2 to dissolve. Calculate the new pH that would result from this addition or srong aecid was aded to his solution in parts(Ce and (

Explanation / Answer

Let us consider the reactions:

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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