# Titrate sulphuirc acid with sodium hydroxide

Coursework time has hit me at college and no matter how much I hit the books the answer is not hitting me.

It might be simple but I can't see it. Of course, the question is needed.

I had to titrate sulphuirc acid with sodium hydroxide. Not hard to write coursework on and I am sure that if I did it without what I am trying to find out (below) then it would be fine but I am not happy just to accept that there should be equal amounts of sulphuric acid and sodium hydroxide to create a neutral solution.

I read in these forums that the molar pH of sulphuirc acid is less than that of the pOH of the sodium hydroxide. For my hypothesis I want to be able to say why I think that there needs to be more sodium hydroixde to create a neutral pH overall.

For what I need help with is the maths, really. I have been studying for the last 4 hours trying to make sense of the dissonance(?) of acids and bases and how this affects the pH or the moles (not even sure which). I installed a system that was posted up here from the 'few questions' thread (called BATE) and it says that for an equal amount of $$0.01$$ $$mol$$ $$dm^{-3}$$ concentrations of each susbstance I should get an overall pH of 6.17. I am trying to find out how to get to this number and how I can explain it along the way.

Even if this wasn't coursework I would have ask the same question (actually, to be honest I don't need it in my coursework it is more for interest but I really want to understand it).

Cheers P.S. After re-reading it I hope it makes sense. What I want to know is how to work out that the pH after mixing, say, 10cm³ of sodium hydroixde to 10cm³ of sulphuric acid (both concentrated to 0.01M) is going to be 6.17, and therefore how much sodium hydroxide is actually needed. Also I will need to explain why so if you know any websites that will do that, to save you writing it, then I would be grateful.

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GCT
Homework Helper
The pH of an equimolar sodium hydroxide and sulfuric acid is 7, both are considered strong acid/base; the product salt of the two is weak with respect acid or base and thus the pH will remain 7. In the case of a different combination of where one of them is a weak acid/base and the other a strong base/acid with equimolar values of both, you'll need to consider that the product salt component of the weak acid/base will actually influence the pH upon the complete neutralization.

Yes, the strong base acid/base will react to completion with the weak acid/base (that is completely neutralized)...the sole factor which affects the pH afterwards is the product salt.

I have done the titration and the results show that for 10.0cm³ of 0.01M sulphuric acid, approximately 11.0cm³ of 0.01M sodium hydroixde is needed. Are you trying to say that on of the substances is strong and the other is weak because with equal concentrations I can't see that making any difference.

I have been looking around and I found these sorts of equations:

$$K_w = [H^+][OH^-]$$

$$K_a = \frac{[H^+][HA^-]}{[H_2 A]}$$ coming from $$H_2A \rightarrow H^+ + HA^-$$

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Borek
Mentor
GeneralChemTutor said:
The pH of an equimolar sodium hydroxide and sulfuric acid is 7

It is not. Equimolar means 'Having an equal number of moles'. pH will be much lower, depending on the H+ concentration.

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BATE - pH calculations, titration curves, hydrolisis

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Borek
Mentor
The Bob said:
I installed a system that was posted up here from the 'few questions' thread (called BATE) and it says that for an equal amount of $$0.01$$ $$mol$$ $$dm^{-3}$$ concentrations of each susbstance I should get an overall pH of 6.17. I am trying to find out how to get to this number and how I can explain it along the way.

The problem is, you will not get this number, because it is wrong - and I have no idea how did you get it using BATE :( Can you give some more details?

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis

Last edited:
GCT
Homework Helper
It is not. Equimolar means 'Having an equal number of moles'. pH will be much lower, depending on the H+ concentration.

ok, first of all, as long as there is a strong acid/base component present all of the base/acid component will be neutralized (think, la chatelier). In all cases of titrations, assuming equimolar reaction, the final pH is actually pertains to the salt product obtained from the acid base reaction.

If you're referring to to the second ionization of sulfuric acid, it is negligible compared to the first. Look up the ionization constants yourself.

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Gokul43201
Staff Emeritus
Gold Member
The pH comes from the concentration of the acid/base as well as the dissociation constant (or you can use Ka/Kb). If the dissociation constants are very close to 1 (Ka/Kb large), the pH is roughly a function of concentration alone.

Borek
Mentor
GeneralChemTutor said:
In all cases of titrations, assuming equimolar reaction, the final pH is actually pertains to the salt product obtained from the acid base reaction.

pH of endpoint depends on the salt present at endpoint regardless of whether the reaction is equimolar or is not.

If you're referring to to the second ionization of sulfuric acid, it is negligible compared to the first. Look up the ionization constants yourself.

I remember them so I don't have to check. pKa1=-3, pKa2 = 2. Difference is large but pKa2 is far from being negligible. NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis

Borek said:
The problem is, you will not get this number, because it is wrong - and I have no idea how did you get it using BATE :( Can you give some more details?
I tried it again and I didn't get the same answer. To be honest because I do not understand about the dissonance of the acids/bases I can't use the system properly.

Borek said:
I remember them so I don't have to check. pKa1=-3, pKa2 = 2. Difference is large but pKa2 is far from being negligible. NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.
This is what I want to know. How to work it out and how to apply it.

Cheers.

GCT
Homework Helper
pH of endpoint depends on the salt present at endpoint regardless of whether the reaction is equimolar or is not.

If it is equimolar than it will depend on the salt alone, if not we'll need to account for the excess of the acid/base.

NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.

I never said that a NaHS04 solution, an acidic solution, was neutral. The pH of HSO4- is not 2.21. The pH due to the first acid dissociation alone is 2.

Difference is large but pKa2 is far from being negligible.

An acid/base classified as strong is deemed for the most part to dissociate almost to completion in water, in this case $K_a$ will be very large (Gokul...way above 1) since $K_a= \frac{[H^+][anion]}{[H_SO_4]}$. $K_{a2}$ is actually $1.2 x 10^{-2}$. There's really no comparison.

Bob, solve this equation for x, assuming a original solution of sulfuric acid of .01M and sodium hydroxide .01M...

$$.012= \frac{[x]^2}{[.01M-x]}$$

Subtract from pH 7. What you can also do is to actually calculate the pH of sulfuric acid as a whole, including Ka for both its acids. After finding the total concentration of H+ produced, find the excess of the latter by subtracting from the concentration of hydroxide. This would be more accurate.

$$.012= \frac{[.01+x][x]}{[.01-x]}$$, solve for x. $[H_3O^+]=[x+.01]$.

I'm sure you can do the rest.

You'll find that the pH of an equimolar solution of sodium hydroxide and sulfuric acid is roughtly 6.9, perhaps a bit higher.

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GeneralChemTutor said:
I'm sure you can do the rest.
Wouldn't be too sure about that. I am studying this one my own and this is the only place I can get help.

So how do you know that $$K _{a2}$$ is 0.012 and why is it that you are using $$K _{a2}$$???

Borek
Mentor
GeneralChemTutor said:
I never said that a NaHS04 solution, an acidic solution, was neutral. The pH of HSO4- is not 2.21. The pH due to the first acid dissociation alone is 2.

On 03-06-2005 at 11:27 you stated:

The pH of an equimolar sodium hydroxide and sulfuric acid is 7

Now you deny it. Further discussion makes no sense.

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BATE - pH calculations, titration curves, hydrolisis

Here is the problem. I understand nothing at all to do with working out pHs or pOHs. I only found out yesterday that there was a unit pOH.

The problem is I have 10 cm3 of Sodium Hydroxide and 10 cm3 of Sulphuric Acid. I mix them together and they, obviously, create an overall pH. I need to work out what that pH is. How do I do it, using the formulas that are needed and an explaination to what each bit is for and means, please.

I would be really, really grateful if someone could explain it to be and, more or less, drag me through it.

Thanks. Borek
Mentor
The Bob said:
This is what I want to know. How to work it out and how to apply it.

Dissociation equation:

$$AcH \leftrightarrow Ac^- + H^+$$

Equilibrium constant:

$$K_a = \frac{[Ac^-][H^+]}{[AcH]}$$

If you know total concentration - say it is $$C_a$$ - you may calculate $$H^+$$ concentration assuming - in accordance with reaction equation - that $$[Ac^-] = [H^+]$$. If so, undissociated acid concentration is $$[AcH] = C_a - [H^+]$$ and

$$K_a = \frac{[H^+]^2}{C_a - [H^+]}$$

which is a quadratic equation. Solve it for $$[H^+]$$:

$$[H^+] = \frac{-K_a + \sqrt{K_a^2 + 4 K_a C_a}}{2}$$

and you know how to perform pH calculation of the weak acid solution.

Assumption that $$[Ac^-] = [H^+]$$ is not always valid - to be precise you should take account of $$[H^+]$$ ions from the water autodissociation. For not very weak acids and not very diluted solutions assumption holds.

Now get back to the sulphuric acid and its equimolar solution with sodium hydroxide - which is the same as $$NaHSO_4$$ solution. You may treat $$HSO_4^-$$ the same way acetic acid was treated above - first dissociation step was already 'consumed' by neutralization and is not influencing the situation (that's not always the case - here is).

As I already wrote $$pK_{a2}$$ equals 2 - so $$K_{a2}$$ (which describes second dissociation equilibrium) equals 0.01. Put it into the equation for the $$[H^+]$$ together with the 0.01 concentration (in the real titration you will have to calculate the dilution factor) and you will get

$$[H^+] = \frac{-0.01 + \sqrt{0.01^2 + 4*0.01*0.01}}{2} = 0.00618$$

and pH = 2.21

Hope that helps.

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BATE - pH calculations, titration curves, hydrolisis

GCT
Homework Helper
Borek, I mentioned a solution of the strong acid base, however you were implying that I had made a statement where an pure acid solution had a pH of 7.

Bob, I'm not able to explain everything about acid/base chemistry right here, what you'll need to do is read up on your own. For now I'll explain a simple case...what is the pH of a solution of .01M sulfuric acid?

We'll assume that the first acid dissociates to completion, this will give .01M of [H+] and [HSO4-] upon dissociation. The pH at this point can be calculated by taking the negative log of the concentration of [H+]....$pH=-log[H^+]$, at this point the pH is 2. However, if you wish to be absolutely exact the you'll need to incorporate the second ionization which has its own dissociation constant $K_{a2}$.

This is where this equation comes in
$$K_{a2}= \frac{[H+][SO_4^{2-}]}{[HSO_4^-]}$$

the initial concentration of [H+] is .01M, 0 for the sulfate, and .01 for the acid. Thus when x M of the acid has dissociated,

$$.012= \frac{[.01+x][x]}{[.01-x]}$$

Solving for x and then finding [.01 + x] will give you the hydronium concentration of which the pH is approximately 1.9.

Borek said:
$$AcH \leftrightarrow Ac^- + H^+$$

Equilibrium constant:

$$K_a = \frac{[Ac^-][H^+]}{[AcH]}$$

If you know total concentration - say it is $$C_a$$ - you may calculate $$H^+$$ concentration assuming - in accordance with reaction equation - that $$[Ac^-] = [H^+]$$
I can accept all of this. This is all fine.

Borek said:
If so, undissociated acid concentration is $$[AcH] = C_a - [H^+]$$ and
$$K_a = \frac{[H^+]^2}{C_a - [H^+]}$$
This is where I get lost. I don't see what you have done and why.

The rest of this post I can, again, accept, so long as if I try another one I can run it through here.

Borek said:
and pH = 2.21
And yes, BATE says that too. Cheers for the help so far.

GeneralChemTutor said:
Bob, I'm not able to explain everything about acid/base chemistry right here, what you'll need to do is read up on your own. For now I'll explain a simple case...what is the pH of a solution of .01M sulfuric acid?
Understood. I am not looking for the easy road. Just the chance to understand.

GeneralChemTutor said:
$$.012= \frac{[.01+x][x]}{[.01-x]}$$
What is the 0.12 from???

GCT
Homework Helper
It is Ka2, the equilibrium constant for the second dissociation of sulfuric acid. So again, the summary is that by neglecting this dissociation, a sulfuric acid solution will be deemed to have a pH of 2; by taking account of the second dissociation, it will roughly have a pH of 1.9. From this you can obtain a good idea of the pH due to equimolar solution pertaining to your initial proposal. For most calculations at most schools/universities, the second dissociation is neglected, so you won't have an opportunity to show off your skills. GeneralChemTutor said:
It is Ka2, the equilibrium constant for the second dissociation of sulfuric acid. So again, the summary is that by neglecting this dissociation, a sulfuric acid solution will be deemed to have a pH of 2; by taking account of the second dissociation, it will roughly have a pH of 1.9. From this you can obtain a good idea of the pH due to equimolar solution pertaining to your initial proposal. For most calculations at most schools/universities, the second dissociation is neglected, so you won't have an opportunity to show off your skills. I am sure I have more questions but I am going to do some independent learning first and see what I understand of this thread and of books.

As I have said I am learning this myself, I mean the teacher haven't set us a task I just want to know, but my teacher is saying I should wait until next year. The problem is if I wait until next year then I will have more to learn rather than just a little revision.

I am going to, now, set myself the task of finding the pH of 10cm³ of 0.03 M Acetic Acid and 10cm³ of 0.03 M of Sodium Hydroxide so be ready for some more explaining of a wrong answer.

Thanks for your help so far. Really do appreaciate it (even if I can't spell it). I assume, GCT, that you have a degree in Chemistry or so, no???

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I got 1.43pH for the mixed solution of Acetic Acid and Sodium Hydroxide.

Borek
Mentor
The Bob said:
This is where I get lost. I don't see what you have done and why.

$$K_a = \frac{[Ac^-][H^+]}{[AcH]}$$

we know that $$[Ac^-] = [H^+]$$ so we substitute:

$$K_a = \frac{[H^+][H^+]}{[AcH]}= \frac{[H^+]^2}{[AcH]}$$

we know that $$C_a = [Ac^-] + [AcH]$$

so

$$[AcH] = C_a - [Ac^-]$$

but $$[Ac^-] = [H^+]$$, so

$$[AcH] = C_a - [H^+]$$

which can be substituted too, yielding:

$$K_a = \frac{[H^+]^2}{C_a - [H^+]}$$

That's all.

Chemical calculators for labs and education
BATE - pH calculations, titration curves, hydrolisis

Borek said:
we know that $$C_a = [Ac^-] + [AcH]$$
How do we know this???

The Bob said:
I am going to, now, set myself the task of finding the pH of 10cm³ of 0.03 M Acetic Acid and 10cm³ of 0.03 M of Sodium Hydroxide so be ready for some more explaining of a wrong answer.
The pH of Acetic Acid is 1.5 because:
pH = -log10([H3O+ / mol dm-3)
so pH = -log10(0.03 mol dm-3) = 1.5 pH

The pOH is going to be 12.5 because:
pH + pOH = 14
so pOH = 14 - pH = 14 - 1.5 = 12.5 pOH

The only difference for the Sodium Hydroxide is that the pH and the pOH numbers are the other way around.

So if I mix them, how do I work out the overall pH??? Is it what you explained earlier???

The Bob said:
The pH of Acetic Acid is 1.5 because:
pH = -log10([H3O+ / mol dm-3)
so pH = -log10(0.03 mol dm-3) = 1.5 pH

The pOH is going to be 12.5 because:
pH + pOH = 14
so pOH = 14 - pH = 14 - 1.5 = 12.5 pOH

The only difference for the Sodium Hydroxide is that the pH and the pOH numbers are the other way around.
$$K_a = \frac{[H_3O^+][A^-]}{[HA]}$$

$$K_a = \frac{[0.03][0.03]}{[?]}$$

This is where I start to get stuck. What is the question mark representing???