Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

I have the answers for the a,b, and e but not c and d :( can anyone help please!
(a) before addition of any KOH____4.09
(b) after addition of 25.0 mL of KOH_____7.53
(c) after addition of 35.0 mL of KOH
(d) after addition of 50.0 mL of KOH
(e) after addition of 60.0 mL of KOH____12.28

Answer 1

HClO constant is 7.55. And thank you!

Answer 2

35.0 mL of 0.210 M KOH

molarity=moles/volume

find moles of OH

do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+

they will cancel out: H+ + OH- -> H2O

but you'll have some left over,
pH=-log[H+]
pOH=-log[OH-]

pH+pOH=14

Answer 3

HOw can I find moles OH?

Answer 4

oh wait, I just set it to M x V = moles right?

Answer 5

yes

Answer 6

And when I do that, I get 7.35, and I do -log (7.35)= -.866 to get pH?

Answer 7

pOH, not pH sorry

Answer 8

you found the moles of OH- but did you subtract moles of H+ from HClO?

Answer 9

would it be 7.55 - 10.5 (from 50 x .21) = -2.95

Answer 10

well you wouldn't have a negative amount of moles.. you'd have H+ minus OH

so 10.5-7.55=2.95

then pH=-log[H+] because you have an excess of H+

Answer 11

Oh, ok thank you!

Answer 12

Is it ok that I get a negative number when i do -log (2.95)? I get -0.469

Answer 13

sorry i forgot to mention you have to convert back to concentration

Answer 14

oh you didn't convert to L initially, did you?

it was supposed to be : 0.21 M x 0.035 L

Answer 15

i;m sorry i should've noticed earlier, anytime you work with molarity, it's in L

Answer 16

Ohh, no I didn't do that earlier. At what point do I go back and cover to Molarity

Answer 17

from the start, everything has to be in liters :S

Answer 18

Mhm, I did that but you said I would have to convert back to concentration. How/at what step am I supposed to do that?

Answer 19

you use the same equation:

molarity = moles/L of solution

except you'd have to add the 2 volumes you mixed.
e.g. 35 mL + 50 mL = 85 mL = 0.085 L

Answer 20

do it at the end when you have excess moles of one species