Acid/base

[nacho]

Solved questions = quoted, unsolved remain outside quotes

Seeing that requires twice as much base to neutralise it than does,
would this mean that Acetic Acid requires 4 times as much, given it has 4 hydrogens? Or is it that only the Hydrogen bonded to the O at the end is able to be donated, so therefore it requires the same amount of base to be neutralised?
thanks.

Edit: Another question.
In a particular titration, methyl red is used instead iof methyl orange of phenolphthalein. What possible reason could be behind this?
The titration is strong base vs strong acid
I know roughly why this is so, but im having difficulty answering it.
The titration is between a strong acid and strong base, so the end point will be around the slightly basic to slightly acidic range, i.e. pH of 5 - 8. Methyl red detects anything between a pH of 4.2 - 6.3 whereas methyl orange and phenolphthalein indicate between 3.1- 4.4 and 8.3 - 10.0 respectively.
Any ideas on how to improve my answer?

3rd question:
Suppose you perform an acid base titraion in which a strong acid is delivered by pipette and a weak base is delivered by burette, and a pH meter is used to follow the reaction. What should be the pH of the reaction mixture at equivalence point?
it should be 7, right, not below 7?

4th question
Suppose the ring etched onto a 20mL pipette is incorrectly placed and, as a result, the pipette delivers 20.10mL when used correctly. If this pipette is used to deliver a base into the reaction flask and the acid which is to be standardised is placed into the burette, what effect will this instrumental error have on the experimental value of the concentration of the acid compared with the true result?

Shouldn't the answer be that the resulting concentration will be lower than the true result?
As we have delivered more NaOH than 20mL, we will require more acid to neutralise it and so more acid needed means a weaker concentration?

Thanks again.

NEW UNSOLVED QUESTIONS:
Predict the effect of the following on the experimental value for the concentration of nitrogein in the lawn feed: would it cause it to be higher than or lower than the true value, or have no effect? Justify each answer.
-Just a headsup : This is a back titration. Reacted lawn feed solution with excess NaOH. Remaining NaOH was reacted against HCl.
Q1. The supplied value for [HCl] was lower than its true value.

Q2. The distilled water was found to have a pH of 6.0

[Water]

Only the H at the OH is donated, and that it is a weak acid, because , it is only partially ionized in water, I believe.

[nacho]

bump, another question.

[Water]

4th question
Suppose the ring etched onto a 20mL pipette is incorrectly placed and, as a result, the pipette delivers 20.10mL when used correctly. If this pipette is used to deliver a base into the reaction flask and the acid which is to be standardised is placed into the burette, what effect will this instrumental error have on the experimental value of the concentration of the acid compared with the true result?

Base                                   +          Acid
You think it is at 20ml                    Greater Titre because to neutralize 20.10ml





Titre of Acid (Increase, because of 20.10 neutralization) than what should be expected   > n(Acid) Increases = n(Base) Increases than what should be in 20.00ml (Mole Ratios)  > Greater Concentration than expected value as it is divided by 20.00ml.


c(Base) > Stronger Than What it should be










Question 2) I don't know, T_T Sorry

[nacho]

4th question
Suppose the ring etched onto a 20mL pipette is incorrectly placed and, as a result, the pipette delivers 20.10mL when used correctly. If this pipette is used to deliver a base into the reaction flask and the acid which is to be standardised is placed into the burette, what effect will this instrumental error have on the experimental value of the concentration of the acid compared with the true result?

Base                                   +          Acid
You think it is at 20ml                    Greater Titre because to neutralize 20.10ml

Titre of Acid (Increase, because of 20.10 neutralization) than what should be expected   > n(Acid) Increases = n(Base) Increases than what should be in 20.00ml (Mole Ratios)  > Greater Concentration than expected value as it is divided by 20.00ml.

Yea but, if we have a higher titre: c = n/v would result in a lower concentration wouldn't it?
Sorry. i'm a little puzzled here.
the mol of Acid is equal to the mol of base, which we believe is x.
in doing c = n/v, the v = volume of titre, correct?
So we have a stable mol, but our volume is greater?

[Water]

If you have a Strong Base,  You'd need more Acid to Neutralize it.



However in this case, You just had more base.....   So you'd need more acid to neutralize it?

However, we didn't know we had more base, and we had more acid to neutralize it. So we'd think that the base would be stronger, when in reality, it only had more of it inside.

Get me ?

[pi]

Seeing that requires twice as much base to neutralise it than does,
would this mean that Acetic Acid requires 4 times as much, given it has 4 hydrogens? Or is it that only the Hydrogen bonded to the O at the end is able to be donated, so therefore it requires the same amount of base to be neutralised?
thanks.

and require the same amount of base (both monoprotic, regardless of 'strength'), requires double.

[nacho]

If you have a Strong Base,  You'd need more Acid to Neutralize it.



However in this case, You just had more base.....   So you'd need more acid to neutralize it?

However, we didn't know we had more base, and we had more acid to neutralize it. So we'd think that the base would be stronger, when in reality, it only had more of it inside.

Get me ?

If this is the case, then yes. But isn't the question asking for the conc of acid?
Let me get this straight; In this particular question, in real life, would we already know the concentration of the Base, or would we already know the concentration of the Acid? Im leaning towards the former one, because we have to find the concentration of the HCl according to the question?
If this is the case, i understand completely, because eg
we have 2M NaOH, we pipette 20.1ml to the reaction flask.
Assume we get a mean titre of 10ml HCl.
This means that [HCl] = n/v
                              = 0.0402 / 0.01 = 4.02M HCl.


Now suppose we had delivered an actual 20.0ml aliquot of NaOH, we have:
n = c x v = 0.02 x 2 - 0.04mol NaOH
n(HCl) = 0.04 mol. mean titre = 10ml
c = n/v = 0.04 / 0.01 = 4M HCl.
In that case, we have a higher concentration than true result.
Confused:  would the titre of HCl be the same in both circumstances?

I feel like im going wrong over here.. So im assuming it is that we already know the concentration of the Acid, rather than the base.
What confuses me here is that the question asks 'what efect will this instrumental error have on the experimental value of the concentration of the acid compared with the true result?'
There is no effect on the concentration of the acid?!? Why would the concentration of the acid change (given that we know it)

[Water]

Suppose the ring etched onto a 20mL pipette is incorrectly placed and, as a result, the pipette delivers 20.10mL when used correctly. If this pipette is used to deliver a base into the reaction flask and the acid which is to be standardised is placed into the burette, what effect will this instrumental error have on the experimental value of the concentration of the acid compared with the true result?


True sorry, misread the question , again, argh, always, I should just give up doing these types of questions in forum.


Anyways, we have known base because , acid is being standardized.

If you had a volume of Base, and you had known concentration of base. And if you increased the volume of the base, without knowing.

This in effect, would cause acid titre to increase, whilst using the n(Base) in 20.00ml aliquot rather than the n(Base) in 20.10ml . I"d suspect then, that the concentration of acid would be weaker.


Equations to justify. Given there is 20ml Titre, for equivalence point to neutralize all of NaOH


NaOH             +            HCL

4M
20.00ml                      20.00ml



n(OH-)  = 0.08mol


n(HCL) = n(NaOH)
          = 0.08

c(HCL) = 0.08mol / 20ml = 4M



Increase Volume in Pipette to 20.10 ml  > Increase Titre to 20.10 ml

c(HCL) =  0.08/ 20.010ml = 3.98M

[Mao]

3rd question:
Suppose you perform an acid base titraion in which a strong acid is delivered by pipette and a weakb ase is delivered by burette, and a pH meter is used to follow the reaction. What should be the pH of the reaction mixture at equivalence point?
it should be 7, right, not below 7?



Theoretically, it should be seven, because the equivalence point, would measure, the exact amount of base is required to neutralize the acid.

Under real circumstances, PH can hardly ever be 7 though.

No.

Theoretically, the pH at the equivalence point should be under 7.

Remember:

Acid + Base --> Conj.Base + Conj.Acid

For strong acids, the conjugate base is neutral. For strong bases, the conjugate acid is neutral. That is, strong --> neutral conjugate always. Thus in the neutralization of NaOH and HCl for example, the products are both neutral, thus equivalence point is 7.

For weak acids, the conjugate base is not neutral, it is partially basic. Similarly for weak bases, the conjugate acid is weakly acidic.

In a reaction of weak acid and strong base, e.g. CH3COOH + NaOH --> CH3COONa + H2O, CH3COONa is the conjugate base of ethanoic acid, H2O is the conjugate acid of NaOH. NaOH is a strong base, thus its conjugate (H2O) is neutral. CH3COOH is a weak acid, thus its conjugate (CH3COONa) is basic. At the equivalence point, all CH3COOH and NaOH have reacted, leaving only H2O and CH3COONa in the system. Due to the basic nature of CH3COONa, the pH would be greater than 7. That is the equivalence point has a pH > 7.

Apply the same reasoning to strong acid + weak base.

[Water]

Thanks for the explaination Mao . Was also curious, if you have a solution for


In a particular titration, methyl red is used instead iof methyl orange of phenolphthalein. What possible reason could be behind this?
The titration is strong base vs strong acid.


Nacho's previous question.

[Mao]

Thanks for the explaination Mao . Was also curious, if you have a solution for


In a particular titration, methyl red is used instead iof methyl orange of phenolphthalein. What possible reason could be behind this?
The titration is strong base vs strong acid.


Nacho's previous question.

Methyl orange and phenolphthalein give fairly late end points. If this was a titration of dilute strong acid and dilute strong base, these indicators would not be suitable even though the change in pH is sharp. If the titration was between concentrated (say, 0.01M) acids and bases, it wouldn't matter much at all.

[Water]

Sorry, for my noob question but

If this was a titration of dilute strong acid and dilute strong base, these indicators would not be suitable even though the change in pH is sharp


Why would they be unsuitable? Is it because the color change wouldn't be sharp enough?

[Mao]

Sorry, for my noob question but

If this was a titration of dilute strong acid and dilute strong base, these indicators would not be suitable even though the change in pH is sharp


Why would they be unsuitable? Is it because the color change wouldn't be sharp enough?

Say if it was a solution of 0.001M HCl vs 0.001M NaOH, then the starting pH would be 3, the equivalence point is 7, and then a limiting behaviour of pH=11 as more NaOH is added. Would you then suggest we use methyl orange 3.1- 4.4 or phenolphthalein 8.3 - 10.0 given that our titration is between 3 and 11?

Now, say if it was a solution of 0.01M HCl vs 0.01M NaOH, our titration range is now between 2 and 12, we can safely use phenolphthalein and methyl orange.

[Water]

Just abit in conflict here.

Been thinking about it all day, and soz for bugging you about this. But,

With titration 3 - 11

Isn't the 3 still beneath 3.1

and the 11 above 10.0


and the 2 and 12, is similar?

So where do you draw the line on why it isn't suitable. ):