# Acid base reaction pH caluclation

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Hi all,

So we have two solutions of 0.341 mol.dm^-3 ethanoic acid and sodium hydroxide. We are asked to calculate the pH of the solution after 12.50cm^3 of sodium hydroxide has been added to 25.00cm^3 ethanoic acid, given the Ka of ethanoic acid is 1.8x10^-5.

The mark scheme using the Henderson-Hasselbrolch equation, e.g. pH=pKa-log([anion]/[acid]), which is fair enough, I just forgot to use that equation. That gives an answer of 4.7

When I tried to do it I figured that the 12.5cm^3 base would react with 12.5cm^3 acid leaving 12.5^cm3 acid remaining. I then put that into Ka=[H+]^2/[Acid], with 12.5cm^3*0.341 mol.dm^-3 as my value for [Acid], and calculated pH as being 3.6. Could someone explain where my thinking is off for the method I tried?

I also noticed that the first method is independent of concentration, given they are both the same? That seems funny. The mark scheme said when half the sodium hydroxide is added there will be equal concs of acid and acid salt, so in the Henderson-Hasselbrolch equation it becomes log[1], but I don't really understand what that means.

I would ask my teacher normally but it's half term for me right now. This was also a specimen paper, so the mark scheme is... minimalistic. Many thanks in advance!

Aaron

So we have two solutions of 0.341 mol.dm^-3 ethanoic acid and sodium hydroxide. We are asked to calculate the pH of the solution after 12.50cm^3 of sodium hydroxide has been added to 25.00cm^3 ethanoic acid, given the Ka of ethanoic acid is 1.8x10^-5.

The mark scheme using the Henderson-Hasselbrolch equation, e.g. pH=pKa-log([anion]/[acid]), which is fair enough, I just forgot to use that equation. That gives an answer of 4.7

When I tried to do it I figured that the 12.5cm^3 base would react with 12.5cm^3 acid leaving 12.5^cm3 acid remaining. I then put that into Ka=[H+]^2/[Acid], with 12.5cm^3*0.341 mol.dm^-3 as my value for [Acid], and calculated pH as being 3.6. Could someone explain where my thinking is off for the method I tried?

I also noticed that the first method is independent of concentration, given they are both the same? That seems funny. The mark scheme said when half the sodium hydroxide is added there will be equal concs of acid and acid salt, so in the Henderson-Hasselbrolch equation it becomes log[1], but I don't really understand what that means.

I would ask my teacher normally but it's half term for me right now. This was also a specimen paper, so the mark scheme is... minimalistic. Many thanks in advance!

Aaron

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#2

(Original post by

Hi all,

So we have two solutions of 0.341 mol.dm^-3 ethanoic acid and sodium hydroxide. We are asked to calculate the pH of the solution after 12.50cm^3 of sodium hydroxide has been added to 25.00cm^3 ethanoic acid, given the Ka of ethanoic acid is 1.8x10^-5.

The mark scheme using the Henderson-Hasselbrolch equation, e.g. pH=pKa-log([anion]/[acid]), which is fair enough, I just forgot to use that equation. That gives an answer of 4.7

When I tried to do it I figured that the 12.5cm^3 base would react with 12.5cm^3 acid leaving 12.5^cm3 acid remaining. I then put that into Ka=[H+]^2/[Acid], with 12.5cm^3*0.341 mol.dm^-3 as my value for [Acid], and calculated pH as being 3.6. Could someone explain where my thinking is off for the method I tried?

I also noticed that the first method is independent of concentration, given they are both the same? That seems funny. The mark scheme said when half the sodium hydroxide is added there will be equal concs of acid and acid salt, so in the Henderson-Hasselbrolch equation it becomes log[1], but I don't really understand what that means.

I would ask my teacher normally but it's half term for me right now. This was also a specimen paper, so the mark scheme is... minimalistic. Many thanks in advance!

Aaron

**Kwork**)Hi all,

So we have two solutions of 0.341 mol.dm^-3 ethanoic acid and sodium hydroxide. We are asked to calculate the pH of the solution after 12.50cm^3 of sodium hydroxide has been added to 25.00cm^3 ethanoic acid, given the Ka of ethanoic acid is 1.8x10^-5.

The mark scheme using the Henderson-Hasselbrolch equation, e.g. pH=pKa-log([anion]/[acid]), which is fair enough, I just forgot to use that equation. That gives an answer of 4.7

When I tried to do it I figured that the 12.5cm^3 base would react with 12.5cm^3 acid leaving 12.5^cm3 acid remaining. I then put that into Ka=[H+]^2/[Acid], with 12.5cm^3*0.341 mol.dm^-3 as my value for [Acid], and calculated pH as being 3.6. Could someone explain where my thinking is off for the method I tried?

I also noticed that the first method is independent of concentration, given they are both the same? That seems funny. The mark scheme said when half the sodium hydroxide is added there will be equal concs of acid and acid salt, so in the Henderson-Hasselbrolch equation it becomes log[1], but I don't really understand what that means.

I would ask my teacher normally but it's half term for me right now. This was also a specimen paper, so the mark scheme is... minimalistic. Many thanks in advance!

Aaron

Having got that out of the way, let's look at the acid dissociation expression:

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

when the acid concentration = the salt concentration [A-] = [HA] and the expression becomes:

ka = [H+]

take logs and change sign

pKa = pH

In the HH equation (derived from the above):

pKa = pH - log[A-]/[HA]

when [A-] = [HA] then [A-]/[HA]= 1

and log[A-]/[HA]= 0

Hence pKa = pH

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#3

(Original post by

The HH equation is rubbish!

Having got that out of the way, let's look at the acid dissociation expression:

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

when the acid concentration = the salt concentration [A-] = [HA] and the expression becomes:

ka = [H+]

take logs and change sign

pKa = pH

In the HH equation (derived from the above):

pKa = pH - log[A-]/[HA]

when [A-] = [HA] then [A-]/[HA]= 1

and log[A-]/[HA]= 0

Hence pKa = pH

**charco**)The HH equation is rubbish!

Having got that out of the way, let's look at the acid dissociation expression:

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

when the acid concentration = the salt concentration [A-] = [HA] and the expression becomes:

ka = [H+]

take logs and change sign

pKa = pH

In the HH equation (derived from the above):

pKa = pH - log[A-]/[HA]

when [A-] = [HA] then [A-]/[HA]= 1

and log[A-]/[HA]= 0

Hence pKa = pH

Having a go at this myself - please could you explain why [A-] =[HA]? Does this assume that the acid completely dissociates?

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#4

(Original post by

Having a go at this myself - please could you explain why [A-] =[HA]? Does this assume that the acid completely dissociates?

**nchmd1999**)Having a go at this myself - please could you explain why [A-] =[HA]? Does this assume that the acid completely dissociates?

It assumes that the weak acid is undissociated = [HA]

And the salt is 100% dissociated = [A-]

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#5

(Original post by

No.

It assumes that the weak acid is undissociated = [HA]

And the salt is 100% dissociated = [A-]

**charco**)No.

It assumes that the weak acid is undissociated = [HA]

And the salt is 100% dissociated = [A-]

It may help me if you could explain why Kwork's reasoning is incorrect as I followed the same thought process? Sorry to be a pain.

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#6

Write down the Ka expression - Ka = [CH3COO-][H+] / [CH3COOH]

Then you know that [H+] = [CH3COO-] so Ka = [H+]^2/[CH3COOH]

After this point I'm stuck - I tried to calculate moles of acid and then concentration in 37.5cm3 but that got me a PH of 2.69, so...

Then you know that [H+] = [CH3COO-] so Ka = [H+]^2/[CH3COOH]

After this point I'm stuck - I tried to calculate moles of acid and then concentration in 37.5cm3 but that got me a PH of 2.69, so...

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#7

(Original post by

Sorry but I'm having trouble understanding this. If we assume that the acid is undissociated then how is the salt even formed?

It may help me if you could explain why Kwork's reasoning is incorrect as I followed the same thought process? Sorry to be a pain.

**nchmd1999**)Sorry but I'm having trouble understanding this. If we assume that the acid is undissociated then how is the salt even formed?

It may help me if you could explain why Kwork's reasoning is incorrect as I followed the same thought process? Sorry to be a pain.

The acid is only dissociated about 1%

CH

_{3}COOH <==> CH

_{3}COO- + H

^{+}

When anything reacts with the hydrogen ions they are removed from the equilibrium, which responds by making more of them (goes to the forward direction until equilibrim is re-established)

This allows the acid to react with base.

According to the equilibrium law:

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

If you input [HA] concentration as (say) 0.1mol/vol the calculation is easier than using (0.1-x) mol/vol, where x is the moles of conjugate base ions formed.

The ka expression would become:

ka = x

^{2}/(1-x)

which, although it can be solved easily enough, gives the answer to within 0.3% of the approximated expression. Hence the approximation above is good and saves time and effort.

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#8

(Original post by

We assume that the acid is undissociated only from the point of view of approximating the calculation and avoiding quadratic equations.

The acid is only dissociated about 1%

CH

When anything reacts with the hydrogen ions they are removed from the equilibrium, which responds by making more of them (goes to the forward direction until equilibrim is re-established)

This allows the acid to react with base.

According to the equilibrium law:

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

If you input [HA] concentration as (say) 0.1mol/vol the calculation is easier than using (0.1-x) mol/vol, where x is the moles of conjugate base ions formed.

The ka expression would become:

ka = x

which, although it can be solved easily enough, gives the answer to within 0.3% of the approximated expression. Hence the approximation above is good and saves time and effort.

**charco**)We assume that the acid is undissociated only from the point of view of approximating the calculation and avoiding quadratic equations.

The acid is only dissociated about 1%

CH

_{3}COOH <==> CH_{3}COO- + H^{+}When anything reacts with the hydrogen ions they are removed from the equilibrium, which responds by making more of them (goes to the forward direction until equilibrim is re-established)

This allows the acid to react with base.

According to the equilibrium law:

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

If you input [HA] concentration as (say) 0.1mol/vol the calculation is easier than using (0.1-x) mol/vol, where x is the moles of conjugate base ions formed.

The ka expression would become:

ka = x

^{2}/(1-x)which, although it can be solved easily enough, gives the answer to within 0.3% of the approximated expression. Hence the approximation above is good and saves time and effort.

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