[annarchy]

hiya

I have a titration curve of a weak acid against strong base which looks like this:

http://www.chemguide.co.uk/physical/...eqia/wasb1.gif

typical WA/SB titration curve.

Concs of the WA and the SB were both 0.1M.

I have a question here which asks me, if the concentration of both the weak acid and the strong base were changed from 0.1M to 0.01M how would this change

-the starting pH
-the pH at the point of 1/2 equivalence
-the size of the pH change close to the equivalence point



I started by thinking the [OH-] and [H+] would both decrease by the same amount.... and then I'm lost any help appreciated!

[AlbertC]

The starting pH would likely drop/increase by 1 point.
For example, if the starting pH of HCl is pH 1 at 0.1M, changing the concentration to 0.01 M would increase the pH to pH2.
This is due to the fact that pH is calculated using -log(concentration). Try it for yourself!

I would assume that the pH at the point of equivalence ( not that it's not 1/2 equivalance point but called equivalance point only from what I've been taught ) would not change as the definition of equivalence point is the point at which equal amounts of acid and base exists.

The size of the pH change would not change also.


Tip : Sometime when you are stuck or in a not familiar situation, try linking it back to the definition.

I notice that the curve you show me has a fault in it? It should have a buffer region because it's a titration between a Weak acid/base and a Strong base/acid. This would cause a buffer to be form and thus the buffer region. Can anyone guide me on this?


Feel free to correct me if I'm wrong

[annarchy]

Originally Posted by AlbertC

The starting pH would likely drop/increase by 1 point.
For example, if the starting pH of HCl is pH 1 at 0.1M, changing the concentration to 0.01 M would increase the pH to pH2.
This is due to the fact that pH is calculated using -log(concentration). Try it for yourself!

I would assume that the pH at the point of equivalence ( not that it's not 1/2 equivalance point but called equivalance point only from what I've been taught ) would not change as the definition of equivalence point is the point at which equal amounts of acid and base exists.

The size of the pH change would not change also.


Tip : Sometime when you are stuck or in a not familiar situation, try linking it back to the definition.

I notice that the curve you show me has a fault in it? It should have a buffer region because it's a titration between a Weak acid/base and a Strong base/acid. This would cause a buffer to be form and thus the buffer region. Can anyone guide me on this?


Feel free to correct me if I'm wrong

Okay, for the first part I agree with you.
I altered the figures in my graph (a bit tough using Ex-hell lol) and you're right! The 1/2 eq. point changes as well but you've tricked me by mentioning the fact that the definition is still what it is, so I'll include this note along with the new figure from my graph.


As for the fault, I am not entirely sure how to obtain the "buffer region" from a pH curve... Pretty terrible for a chem student but I only know the buffer regions by heart according to commonly used acid and base buffers... Cannot deduce a buffer region from a graph. How do you do this?

thank you for your reply btw!

[AlbertC]

Good point about the equivalence point changes as well. BTW, why are you using the term 1/2 equivalence point, but not just only equivalence point? I thought it should be only "A" point.

This is two wepsite I found by googling that might help you :

http://chemistry.beloit.edu/Rain/pages/titr.html
http://dwb.unl.edu/calculators/activities/titrate.html

From those website, it seems I'm wrong about the last 2 questions you were asking. But another questions arises, how do you calculate equivalence point?

Buffer region is a gradually decreasing portion of the curve before any steep changes in pH to the equivalence point, should be the same for all weak acid-strong base titration curve and weak base-strong acid titration curve.

But I'm not sure if there is any buffer region in strong acid-weak base and strong base-weak acid titration. That is what I wanted to ask. According to theory any titration between weak Acid/Base with strong Acid/Base should have a buffer region. Can anyone clarify on this?

Your welcome!

[charco]

Seems to be some confusion regarding the equivalence point and the 1/2 equivalence point. They are two completely different things.

The equivalence point is when all of the acid/base has been neutralised. It is the amount of titrant that has to be added to completely react with all of the other reactant.

The half equivalence point is a volume that is equal to HALF the volume required to get to the equivalence point.

At the equivalence point the pH is the same as that of the hydrolysed salt.

At the half equivalence point pH = pKa of the acid.

[JakeyTheSnake]

pH = -log[H+]

For the first one, the conc of the OH- is 0.1 originally. We need conc H+, so we use the 10^-14 = [H+][OH-]. Therefore conc [H+] = 10^-14 divided by 0.1. This gives us a very small answer, and we can -log it to get the original pH value, which is 13.

Repeat for 0.01, you get pH 12.

Because [H+] [OH-] = a constant, if [OH-] goes down by a factor of 10, like it has, [H+] goes up by a factor of 10. pH = -log[H+], and you should know logs well enough to know that when your value goes up by a factor of 10, the answer will have a difference of 1 (when in base 10). Then you can deduce logically that the conc of OH- ions is going down, therefore H+ is going up, so it must be getting more acidic...therefore it drops by one down to 12

But if in doubt work it out!

[AlbertC]

Originally Posted by charco

Seems to be some confusion regarding the equivalence point and the 1/2 equivalence point. They are two completely different things.

The equivalence point is when all of the acid/base has been neutralised. It is the amount of titrant that has to be added to completely react with all of the other reactant.

The half equivalence point is a volume that is equal to HALF the volume required to get to the equivalence point.

At the equivalence point the pH is the same as that of the hydrolysed salt.

At the half equivalence point pH = pKa of the acid.

OHH!! haha, I wasn't taught about the half equivalence point. Sorry

BTW, how do I to calculate the equivalence point?
What information do you need?
What does the half equivalence point tells you about?

It would be nice if you could clear the confusion about the buffer region also. Does it occur between weak Acid/base-strong Acid/base titration regardless of whether which came first? THANKS! Sorry to hijack your thread TC!

[charco]

Originally Posted by AlbertC

OHH!! haha, I wasn't taught about the half equivalence point. Sorry

BTW, how do I to calculate the equivalence point?

You can either do it by experiment (titration) or you can use the moles equivalent from the equation.

Eg if you are reacting 0.1 M NaOH with 20cm³ 0.2 M HCl

The equation for the reaction:

HCl + NaOH --> NaCl + H₂O

tells you that the moles of HCl equals the moles of NaOH at the equivalence point.

Moles of HCl = 0.2 + 0.020 = 0.004 mol
Moles of NaOH = 0.004 mol

Molarity of NaOH = 0.1
Therefore volume of NaOH needed = 0.004/0.1 = 0.04 dm³ = 40 cm³

What information do you need?

see above...

What does the half equivalence point tells you about?

The half equivalence point is specially relevant for weak acid/strong base or weak base/strong acid titrations.

For a weak acid:

HA <==> H+ + A-

and

Ka = [H+][A-]/[HA]

at the half equivalence point half of the acid (HA) has been used up and made an equal amount of salt (A-)
So, [HA] = [A-]

substituting that into the Ka equation gives:

Ka = [H+]

Hence pKa = pH

Using this allows you to find the pKa value of a weak acid by titration. (provided that you measure the pH at the half equivalence point) as part of the procedure, or after it in a separate test.

It would be nice if you could clear the confusion about the buffer region also. Does it occur between weak Acid/base-strong Acid/base titration regardless of whether which came first? THANKS! Sorry to hijack your thread TC!

The buffer region occurs when a weak acid is in solution with its salt
OR when a weak base is in solution with its salt.

When this occurs depends on what you are adding to what.

If you add strong base to a weak acid, then a buffer results immediately any of the acid is neutralised, as salt is formed

However, if you add weak acid to strong base you never have any of the weak acid present with the salt until after all of the base is neutralised.

By similar logic if you add weak base to strong acid there will only be weakl base in conjunction with its salt AFTER all of the acid has been used up.

But, if you add strong acid to weak base there will be a buffer formed as soon as any reaction occurs.

[AlbertC]

Oh thanks!
Very detailed explanation indeed charco!

For the first question I was meaning to ask how do you calculate pH of the equivalence point. Sorry for the misleaading question. Nonetheless, I am still deeply appreciated that you went trough all that trouble typing all that out.

My main problem about going calculating the pH of the equivalence point is because I thought that both the acid and base have already completely reacted and form salts. Thus, there shouldn't be any [H+] or [OH-] to give me a calculation. Therefore I arrive at the conclusion that the pH now is due entirely to the salt form. But then, how do you go about calculating the pH of a acidic/basic/neutral salt? Is there any way to it?

[annarchy]

Originally Posted by charco

The buffer region occurs when a weak acid is in solution with its salt
OR when a weak base is in solution with its salt.

When this occurs depends on what you are adding to what.

If you add strong base to a weak acid, then a buffer results immediately any of the acid is neutralised, as salt is formed

However, if you add weak acid to strong base you never have any of the weak acid present with the salt until after all of the base is neutralised.

By similar logic if you add weak base to strong acid there will only be weakl base in conjunction with its salt AFTER all of the acid has been used up.

But, if you add strong acid to weak base there will be a buffer formed as soon as any reaction occurs.

Cool, I've already handed in the work but for interest's sake anyway where is the buffer region on a typical weak acid/strong base titration curve graph?

thanks for this answer btw

[charco]

Originally Posted by annarchy

Cool, I've already handed in the work but for interest's sake anyway where is the buffer region on a typical weak acid/strong base titration curve graph?

thanks for this answer btw

I've explained this in the post above.

[charco]

Originally Posted by AlbertC

Oh thanks!
Very detailed explanation indeed charco!

For the first question I was meaning to ask how do you calculate pH of the equivalence point. Sorry for the misleaading question. Nonetheless, I am still deeply appreciated that you went trough all that trouble typing all that out.

My main problem about going calculating the pH of the equivalence point is because I thought that both the acid and base have already completely reacted and form salts. Thus, there shouldn't be any [H+] or [OH-] to give me a calculation. Therefore I arrive at the conclusion that the pH now is due entirely to the salt form. But then, how do you go about calculating the pH of a acidic/basic/neutral salt? Is there any way to it?

To calculate the pH of a salt you need to know the pKa of the weak acid (or base) that made it.

The salt ions get hydrolysed by water:

A- + H2O <==> HA + OH-

a hydrolysis equilibrium constant can be defined as:

kh = [HA][OH-]/[A-]

and we know that

kw = [H+][OH-], therefore [OH-] = kw/[H+]

so substituting for [OH-]

Kh = [HA]Kw/[A-][H+]

and we know that

Ka = [A-][H+]/[HA]

so that Kh = Kw/Ka

Example:

if the concentration of sodium ethanoate is 0.1M

Kh = Kw/Ka = 1 x 10^-14/1.78 x 10^-5

Kh = 5.62 x 10^-10 = [HA][OH-]/[A-]

we assume that [A-] is almost unchanged (close approximation) and that [HA]=[OH-] (stoichiometry)

[OH-] = root(0.1 x 5.62 x 10^-10) = 7.5 x 10^-6

pOH = -log[OH-] = 5.125

pH = 14 - pOH = 8.87

[AlbertC]

Originally Posted by charco

To calculate the pH of a salt you need to know the pKa of the weak acid (or base) that made it.

The salt ions get hydrolysed by water:

A- + H2O <==> HA + OH-

a hydrolysis equilibrium constant can be defined as:

kh = [HA][OH-]/[A-]

and we know that

kw = [H+][OH-], therefore [OH-] = kw/[H+]

so substituting for [OH-]

Kh = [HA]Kw/[A-][H+]

and we know that

Ka = [A-][H+]/[HA]

so that Kh = Kw/Ka

Example:

if the concentration of sodium ethanoate is 0.1M

Kh = Kw/Ka = 1 x 10^-14/1.78 x 10^-5

Kh = 5.62 x 10^-10 = [HA][OH-]/[A-]

we assume that [A-] is almost unchanged (close approximation) and that [HA]=[OH-] (stoichiometry)

[OH-] = root(0.1 x 5.62 x 10^-10) = 7.5 x 10^-6

pOH = -log[OH-] = 5.125

pH = 14 - pOH = 8.87

Hey, thanks man. Helped me out a lot! Really appreaciated the help!