Question

question involving a triprotic acid

Answer 1

anyone can help me for part c?

Answer 2

i found values for part b to be\[[H _{2}A ^{-}]=0.04\]
\[[HA ^{2-}]=2.5\times10^-6\]
\[[A ^{3-}]=6.25\times10^{-13}\]

Answer 3

\([H_2A^-]\) is ok, but the other values are not correct.

How have you found the value 0.04 ?

Answer 4

oh oh, sry i think i made a mistake, should [H2A-]=0.01 or 0.08?,

Answer 5

No, this value was ok, but the other ones are wrong.

Answer 6

\[\frac{ x^2 }{ 0.2-x }=0.01\] isnt it?

Answer 7

[HA2-]=10^-5
[A3-]=2.5x10^-12
is this correct now?

Answer 8

Now, everything is perfect! Well done!

Answer 9

I'm not sure the last question is well worded.
I would write:
"What is the pH if you add 0.55 mole of strong base to 1 litre of the 0.20 M solution of H3A?"

Answer 10

okay great! :D hm, i attempted the qn too and i got pH 2.23 i think. bt not sure if its correct

Answer 11

There are enough OH- to get rid of the first two protons of the acid.
So the equilibrium will be between \(HA^{2-}\) and \(HA^{2-}\)

The pH will be around 8.

Answer 12

ohh uhm did i draw the reaction table correctly? how do i get pH 8?

Answer 13

No, you have to react H3A with OH- : OH- will be in excess and H2A- will be formed.

Then, you have to react H2A- with OH- : OH- will still be in excess and HA2- will be formed.

Finally, react HA2- with OH- : OH- will be used up, HA2- will be in excess, A3- will be formed.
The ratio between A3- and HA2- will lead to the pH of the solution.

Answer 14

i got 7.15 as my answer is that correct?

Answer 15

I don't think so, but in order to check, what are your final amounts of A3- and HA2- ?

Answer 16

for HA2- it is 0.05 for A3- is 0.35?

Answer 17

This is impossible, since you had 0.20 mol of H3A in the first place.

Answer 18

what must i do to amend?, its still vague to me

Answer 19

I wrote everything here earlier:

1. You have to react H3A with OH- : OH- will be in excess and H2A- will be formed.
2. Then, you have to react H2A- with OH- : OH- will still be in excess and HA2- will be formed.
3. Finally, react HA2- with OH- : OH- will be used up, HA2- will be in excess, A3- will be formed. The ratio between A3- and HA2- will lead to the pH of the solution.

Can you set up a table corresponding to step 1. ?

Answer 20

the first one H3A reacted with OH- i got 0.35 OH- left, how much more would H2A- form? wouldnt it be 0.2 also? or i must add to make it 0.75?

Answer 21

is this correct? for the first one