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
5 answers
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.
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.
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.
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.
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.
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.
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.
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.
1 answer