100 mL of a .256 mol/L Acetic acid solution is added to 150.00 mL of a .256 mol/L magnesium acetate. The magnesium acetate solution is also 0.0100 mol/L magnesium chloride.
a) Write a balanced equation for the dissolution of each compound in water.
b) Write the balanced chemical equilibrium/equilibria that exist in this solution.
c) Calculate the concentration of all chemical species in the solution.
d) Calculate the pH of the resulting solution.
Any help with problem will help!! Thank you!!!
I don't know if this is a freshan level course or an upper class problem. It seems too much for beginning class and posed in a way that isn't clear for upper class. I'm thinking it is a beginning class but the author of the problem has failed to take into account all of the intricacies but here goes. Some of the interactions I'll just ignore. Here is a summary of what you have.
HAc = acetic acid = 100 mL x 0.256 M = 25.6 millimoles.
Mg(Ac)2 = magnesium acetate = 150 mL x 0.256 M = 30.4 millimoles
MgCl2 = magnesium chloride = 150 mL x 0.01 M = 1.5 millimoles
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HAc + H2O ==> H3O^+ Ac^-
Mg(Ac)2 + H2O ==> Mg^2+(aq) + 2Ac^-(aq). Usually the solution of the salt is written with H2O placed over the arrow so show water is there and the (aq) takes care of it on the right side but I can't place the H2O there. The others are done the same way.
MgCl2 + H2O ==> Mg^2+(aq) + 2Cl^-(aq)b2. I have omitted the (aq) for hereafter; you can add them if you wish.
b. HAc + H2O ==> H3O^+ Ac^-
H2O ==> H^+ + OH^-
Ac^- + H2O ==> HAc + OH^-
c. I have a problem with the way the problem is stated. Calculation of (HAc) and the others is relatively simple IF YOU ASSUME JUST THE ONE MATERIAL IS THERE AND THERE ARE NO INTERACTIONS. Adding Ac^- for Mg(Ac)2 changes that. I will assume you want concn with the interactions although I may ignore some of less obvious ones . First for HAc. You have 100 + 150 mL = 250 mL in the solution and HAc is 25.6 mmols so (HAc) initially is M = millimoles/mL = 25.6/250 = 0.102 M and that is in a solution of (Ac^-) = (2*30.4)/250 mL = 0.243 M
.....................HAc + H2O ==> H3O^+ Ac^-
I....................0.102...................0............0.243
C......................-x......................x.............x
E.................0.102-x...................x.............0.243+x
Plug the E line into the Ka expression for HAc and solve for H3O^+ and Ac^-.
For the salts Mg(Ac)2 and MgCl2.
(Mg^2+) = mmols Mg from Mg(Ac)2 + mmols Mg from MgCl2 (find those mmoles in the summary at the top) divided by total volume in mL
For (Ac^-) that will be twice the concn of Mg^2+ in the Mg(Ac)2 and (Cl^-) will be twice concn of Mg^2+ in the MgCl2. Note that I did NOT add Ac^- from the HAc ionization to the Ac^- already there. If it makes a significant difference you can do that. Also, the hydrolysis of the Ac^- diminishes Ac^- slightly but I didn't do that either.
d. pH of the solution. I will assume here that the pH of the solution is determined by the buffer solution of HAc and Mg(Ac)2. You do that one by using the Henderson-Hasselbalch equation which is
pH = pKa + log (base)/(acid)
pKa is the pKa for Ka of HAc. pKa = -log Ka. I think pKa for HAc is 4.74 to 4.76 but it may differ slightly depending upon the exact value you use for Ka for HAc. It would be interesting to compare the pH found by the HH equation and that from the first calculation of H^+ at the beginning.
Post your work if you get stuck.
(base) = concn of the acetate
(acid) = concn of the acid HAc