# chem

posted by .

Calculate the pH of the solution that results from mixing the following four aqueous solutions together:

(1.) 100. mL of 0.300 M HI
(2.) 200. mL of 0.250 M LiCl
(3.) 300. mL of 0.200 M KOH
and
(4.) 150. mL of 0.150 M HBr

a. ) 2.00
b. ) 3.99
c. ) 5.32
d. ) 11.88
e. ) 12.00

• chem -

The question is a little confusing (actually very confusing). I assumed on the first reading that you had all four solutions mixed together and you wanted the pH of the one (1) final solution. On the second reading I see you have five answers for four solutions. So I don't get it. If answer a is for the solution of 1 and 4, then I don't think answer a is correct. If answer b is for the mixture of 2 and 4, I don't get that answer either. Please clarify the question.

• chem -

your first assumption was correct. you have to find the pH fo the final solution. the final pH will be one of the 5 options provided.

• chem -

You need to recognize what is a strong acid, strong base, weak acid, weak base, etc.
moles 1 = M x L. Strong acid.
moles 2 = 0 (LiCl is the salt of a strong acid/strong base so it will be neutral--howeer, the volume will count as a dilution).
moles 3 = M x L. Strong base.
moles 4 = M x L. Strong acid.
Total moles strong acid = ??
less total moles strong base = ??
If I didn't goof you will have an excess of the base. Then (OH^-) = moles/total volume in liters.
pOH = -log(OH^-) and pH = 14 - pOH.
I got one of the answers but I think the problem has all of the answers for almost any miscue that we could make.

• chem -

I got an excess of acid.

I added the moles of H+ from HI with the the moles from HBR (0.03+0.0225 = 0.0525mol)

I got 0.04 moles of OH-
Subtacting them I got 0.0125 moles
pH = -log0.0125 = 1.9

I'm not sure if what I did is correct.

• chem -

You should have obtained an excess of base.
100 mL* 0.3 M HCl = 30 mmoles HCl
forget LiCl for the moment; it is neutral.
300 mL*0.2 M KOH = 60 mmoles KOH.
150 mL*0.15 M HBr = 22.5 mmoles HBr.

30 mmoles HCl + 22.5 mmoles HBr = 52.5 mmoles total strong acid.
60 mmoles KOH.
Excess KOH by 60-52.5 = 7.5 mmoles.
(OH^-) = KOH/mL = 7.5 mmoles/750 mL = 0.01 M = (OH^-).
So pOH = 2
pH = 12.
Total volume is 100 + 200 + 300 + 150 = 750 mL = 0.750 L.
I used millimoles (mL x M = mmoles) but I told you to use M x L = moles. You should get the same answer of 0.01 M for OH but not have all of the zeros. For example, 100 mL x 0.3 M = 30 mmoles but 0.1 L x 0.3 M = 0.03 moles).
I hope this helps. Sorry I went to bed on you last night; I assumed you could finish both of the problems.