Acid - Base Reaction - How to calculate the resulting pH?

Question

My daughter is doing an experiment where she knows the pH of various solutions. She adds an acid with a base and were expecting the resulting solution to be pH neutral.

The vinegar - baking soda reaction turned out to be pH 10, no matter what percentage of baking soda was added to the vinegar.

I thought the formula would simply be (pH1 + pH2)/2, so we expected the following:

Vinegar - pH = 2
Baking Soda - pH = 10

Resulting solution (2+10)/2 = 6

Not what we measured. Any ideas?

So what you are saying is that I put too much Baking Soda into the Vinegar? How do I know when I've put in enough to only neutralize it?

Is the general formula we were using valid? (pH1+pH2)/2?

Answer 1

on adding vinegar (typically 5% acetic acid), to NaHCO3 (baking soda) in small quantities. fizzing occurs(effervescence) and CO2 is liberated. At neutralisation point there is no fizzing on adding extra acid to soda. all u need 2 do is to add the acid in small quantities at a time until the mixture no longer fizzes.
the formula u are using is wrong.
if u read on u will notice i mentioned something about equivalence point this reaction has none. therefore no indicator will give a sharp end point so we are back with the same problem of adding excess baking soda.


how to calc pH
NaHCO3 + CH3COOH → CH3CO2Na + H2O + CO2 (gas)
in the titration(note that this is pH titration not volumetric) of weak acid and a weak base a titration curve with no equivalence point is obtained so pH at endpoint is difficult to determine, the way out is to titrate the vinegar against a strong base and the baking soda against a strong acid .you would only get pH7 if u were using a strong acid (HCl) and a strong base (NaOH). if u were doing a titration of weak acid(acetic acid) and a strong base (NaOH) then the resulting pH is easy to determine.
VINEGAR AND STRONG BASE
the experiment - fill burette with NaOH. pipette 25cm3 of acid into conical flask. add 2cm3 of base at a time and use a pH meter to measure pH at each time. plot graph of pH against volume of NaOH.you will get a graph with a vertical region this region is called the equivalence point find mid-point of this region .then from that point get the pH.
BAKING SODA AND STRONG ACID
same as above the baking soda goes in the burette.
I'm surprised u got pH10 I think what happened was that u added more than enough baking soda that was required to neutralise the acid so the resulting pH u got is now of the excess baking soda (unreacted)
THERE A DIFFICULT CALC. U COULD DO USING THE pkb value of baking soda, 6.3. BUT LIKE THE FAILED EXPERIMENT THE RESULTING PH IS DIFFICULT TO OBTAIN.
YOUR DAUGHTER SHOULD CONSIDER DOING A SIMPLER EXPT AND CALC. THE RESULTING pH USING THE APPROACH ABOVE.
NOTE-vinegar is typically 5% ethanoic (acetic) acid. conc of 5% is about 0.833mol/dm3

Answer 2

HAc + NaHCO3 → Ac- + Na+ + H2O + CO2^

Ac- + Na+ + H20 <------> NaOH + HAc

In the latter reaction, NaOH is a strong base but HAc is a weak acid so the result will always be basic if all of the HAc is neutralized. If you have added excess NaHCO3 the solution becomes a buffer which has a pH resistance to changes in the salt's concentration.

EDIT: in answer to your added comment, the crude way is to add small quantities until no fizzing occurs. That will not be accurate, since once fizzing stops you have gone too far. To get a neutral solution, you need to use an acid/base indicator such as litmus or phenolphtalien. You can get pH indicator strips (which change color according to pH) from swimming pool supply stores.

When you end up with a truly neutral (pH 7) solution, not all of the vinegar is reacted.

No, the formula is not valid. The actual computation is more complicated and involves equilibrium constants (acid dissociation constant) for the weak acid. The amount of bicarbonate needed can be calculated from the pH numbers you gave:

The approach is to calculate the concentrations of both the acid and the base using the pH numbers and from that compute how much of each component is needed.

Answer 3

Ok, it's been a while, but I think this may be why:

It may remain slightly basic because NaCH3COO (sodium acetate) which is the salt that forms from the reaction will ionize to form Na+ and CH3COO-. Since NaOH is a strong base, the conjugate acid Na+ will be very very weak and can be neglected. Since CH3COOH is a weak acid, the conjugate base CH3COO- will be a weak base.

When the CH3COO- reacts with H20 it will form CH3COOH and OH-, raising the pH of the solution. The concentration of the acetate ions and thus the concentration of vinegar will therefore affect the final pH, so changing the concentration of baking soda shouldn't matter too much because we can neglect the impact of the sodium ions.

To test this, try changing up the concentration of vinegar you use rather than the amount of baking soda and see if the final pH changes at all.

Answer 4

My chemistry coaching is that NaOH reacts with CH3COOH in a a million:a million mol ratio: Equation NaOH + CH3COOH = CH3COONa + H2O I relatively have calculated which you have 0.0.5 mol CH3COOH and in basic terms 0.0.5 mol of NaOH . this suggests that the CH3COOH is in molar extra whilst in comparison with the NaOH . you have a answer containing the reaction product CH3COONa and the unreacted CH3COOH. in accordance to what I relatively have learnt this could be a buffer answer . The pH is labored out using the Henderson equation. The NaOH extra would not react in basic terms with the dissociated H+ of the acid yet with the acid in a a million:a million molar ratio. i think of that that's a hardship-loose way of bobbing up an ethanoic acid / sodium ethanoate buffer via including NaOH to a answer of the acid as long because of the fact the acid is in molar far extra suitable than the backside. that's what you have right here.

Answer 5

As we dont have concentrations and amount of acid and base.it is better to conduct the experiment with the help of pH meter,add acid into the base solution taken in a beaker in which pH electrode is already immersed and find out pH from the meter reading,
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