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Posted: 13 years ago

Hey, thanks for your help. I just have one more question. When finding the [H+] for the above two points, do we still assume pH=pKa and Ka = [H3O+] or would it be different as the points are before and after half equivalence point? Edited by Miggi - 13 years ago

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

pH and [H3O+] values change but pKa and Ka values do not change.

pKa = pH at half titration point

So at all points,

pKa = 4.1

and Ka = 10^-4.1

Ka = 7.9433 * 10^-5

You already know pH values. So find [H3O+] using the formula

[H3O+] = 10^-pH

At one point, pH = 3.7

Therefore [H3O+] = 10^-3.7 = 2.00 * 10^-4

At the second point, pH = 4.5

Therefore [H3O+] = 10^-4.5 = 3.16 * 10^-5

Edited by akhl - 13 years ago

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143708

Inactive

Posted: 13 years ago

Thank you so much for your help.😊

for the first question ,I got [A-] = 0.0296M is that right?

and in the last question it says to find out

Ka for those two points too. So,shouldn't the Ka be different?

I used the formula [H3O+]=Ka[HA]/[A-] and got Ka of 9.212*10^-5 for first one and 7.804*10^-5 for

2nd point. So, does that mean it's wrong or is it right?

Edited by Miggi - 13 years ago

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

You have to calculate Ka values at both points. But both should be roughly equal, otherwise something wrong in collecting the data or reading the graph.

Ignore what I wrote earlier. In this I am giving you complete answer using ICE table.

MW of ascorbic acid = 176.13

Moles of HA added = 0.8994/176.13 = 0.005106

At the first point:

Before titration:-
Total volume = 50 ml + 4.00 ml = 54 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 4.00 ml/54 ml
[NaOH] = 0.02963 M

Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (4.00 * 10^-3)
Moles of HA = 0.003506
[HA] = 0.003506 moles/(54 ml) = 0.003506/(54*10^-3)
[HA] = 0.0649 M

HA --------> H+ + A-
0.0649          0         0      (I)
-x               +x        +x    (C)
0.0649-x       x         x       (E)

x = [H+] = 10^-pH = 10^-3.7 = 2.00 * 10^-4 = 0.0002

[A-] = x + [NaOH] = 0.0002 + 0.02963 = 0.02983 M

[HA] = 0.0649 - x = 0.0649 - 0.0002 = 0.0647 M

Ka = [H+] [A-]/[HA] = 0.0002 * 0.02983/0.0647 = 9.2 * 10^-5

At the second point:

Before titration:-
Total volume = 50 ml + 9.00 ml = 59 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 9.00 ml/59 ml
[NaOH] = 0.06102 M

Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (9.00 * 10^-3)
Moles of HA = 0.001506
[HA] = 0.001506 moles/(59 ml) = 0.001506/(59*10^-3)
[HA] = 0.0255 M

HA ---------> H+ + A-
0.0255          0         0      (I)
-x               +x        +x    (C)
0.0255-x         x         x       (E)

x = 10^-pH = 10^-4.5 = 3.1623 * 10^-5 M

[A-] = x + [NaOH] = 3.1623 * 10^-5 + 0.06102 = 0.0611 M

[HA] = 0.0255 - x = 0.0255 - 3.1623 * 10^-5 = 0.02547 M
Ka = [H+] [A-]/[HA] = (3.1623 * 10^-5) (0.0611) / 0.02547 = 7.59 * 10^-5

Edited by akhl - 13 years ago

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

See the above solution. There is some difference between the two Ka values. This means there is some error in taking the data.

But it is ok. Some small error will always be there. Just confirm once more that the data are correct.

Edited by akhl - 13 years ago

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143708

Inactive

Posted: 13 years ago

hey thank you so much and i have got the same answers as yours by the different method but i have written down this method too.

Thanks again :)

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143708

Inactive

Posted: 13 years ago

Thanks in advance:)

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

Show the same H atom bonded to both N atoms. It will be covalently bonded to one N atom (solid line) and hydrogen bonded to another N atom (dashed line). Draw as shown below:-

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

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akhl

@akhl

IF-Rockerz

Posted: 13 years ago

Answer to the last problem:

In the last problem you need three compounds. I am giving their structures below.

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