Question

"Solve the following problems using the titration date given"

Answer 1

(A) "How many mL of 0.20 M NaOH are required to neutralize completely...
...8.5 mL of 0.50 M \(\large{H_{2}SO_{4}}\)?\[H_{2}SO_{4}+2NaOH→Na_{2}SO_{4}+2H_{2}O\]

(B) "A 10.0 mL sample of \(\large{H_{3}PO_{4}}\) is placed in a flask....
....Titration required 42.0 mL of 0.10 M NaOH. What is the molarity of the \(\large{H_{3}PO_{4}}\)?"\[H_{3}PO_{4}+3NaOH→Na_{3}PO_{4}+3H_{2}O\]

Answer 2

Do I use the "M1V1=M2V2" equivalency?

Answer 3

Ok so starting with A.
I think it is possible that we can use \[M_1V_1=M_2V_2\] we just have to consider that H2SO4 gives off 2 H+ per molecule while NaOH only gives off 1 OH-.
So say we can say \[2\times(.085L)(.5M)=.2M(V_2)\]. tbh not 100% sure this is legal way of doing this.

Answer 4

so V_2 = .425 L of NaOH

Answer 5

Oh, I see. o:

Answer 6

Oh i just realized that H2SO4 being a polyproptic acid doesn't fully give off its second hydrogen ion. Let me search real quick to make sure im not giving you bad info for calculations.

Answer 7

Okay np

Answer 8

Ok, we did it correctly. Given that we are not given the \(K_a\) for the second proton that would be the correct calculation to do above.

Answer 9

What's the \(\large{K_{a}}\)?

Answer 10

whew glitches are bad on this side x_x

Answer 11

Ok now for question B.
We assume that \(H_3PO_4\) gives off 3 protons and one again NaOH can only give off one hydroxide.
So we say \(3\times(.010L)(M_1)=(.042L)(.10M)\) and then solve for \(M_1\) which comes to be \[M_1=1.4M\]

Answer 12

for some reason text keeps disappearing for me making it damn near impossible to type correctly lol

Answer 13

for me the typing thing remains after I submit Post and sometimes my icon disappears from "viewing" ;-;

Answer 14

and the \(K_a\) represents The acid dissociation constant and looks like this
\[K_a = \frac{ [X^-][H_3O^+] }{[HX][H_2O] }\] HX being the original example acid and X- being the conjugate base.

Answer 15

I... have never heard of that.
Sorry D:

Answer 16

Ok then definitely don't have to worry about it in titration questions :P.

Answer 17

Alright lol

Answer 18

In terms of polyproptic acids basically the 2nd Hydrogen ion will have a lower K_a value (meaning the 2nd hydrogen ion is less likely to ionize into solution). And if the compound/molecule has a third hydrogen to give off its K_a will tend to be even lower then the 2nd hydrogen's K_a value.

Answer 19

Ah... okay? o_o

Answer 20

Uh if thats not helpful and just confusing just ignore it. I was just trying to elaborate on K_a.

Answer 21

No it's okay, I'm just trying to multitask and rid myself of these glitches hahah

Answer 22

The glitches are killer on my end too. Sometimes makes using this site an absolute pain.

Answer 23

agreed

Answer 24

Ill try to be online for a bit. Just tag me if you have some questions. I'll try to help if I can.