What Volume of Ethanoic Acid is Needed to Neutralize 25ml of 0.1 M NaOH?

[grey]

Hi,

I came across a problem, which I might have solved a few years ago when I had studied Chemistry, but now I am totally stumped. Would appreciate some quick solutions.

I have 25 ml of 0.1 M NaOH, and a volume of 0.1 M of Ethanoic Acid. What volume of ethanoic acid would neutralize the 25 ml of the base?

I have a hand on this, but my solution is wrong because it gives 25 ml of ethanoic acid, which shouldn't be the case and that the volume must be greater than 25 ml.

Thanks in advance!

 

[SpectraCat]

Hi,

I came across a problem, which I might have solved a few years ago when I had studied Chemistry, but now I am totally stumped. Would appreciate some quick solutions.

I have 25 ml of 0.1 M NaOH, and a volume of 0.1 M of Ethanoic Acid. What volume of ethanoic acid would neutralize the 25 ml of the base?

I have a hand on this, but my solution is wrong because it gives 25 ml of ethanoic acid, which shouldn't be the case and that the volume must be greater than 25 ml.

Thanks in advance!

It is greater, but only by a very small amount. Assuming you haven't made any typos above, you will have reached the equivalence point after adding your 25 mL of 0.1 M acetic acid. You can then consider that you have a 0.05 M solution of sodium acetate. The pH of that solution is already pretty close to neutral, but if you really need to know how much more you need to add to get it to neutrality (pH=7), you can use the Henderson-Hasselbach equation to calculate the final concentrations of acetic acid and acetate that you need for your final solution. Since the buffering region of acetic acid/acetate mixtures is between pH=3.75 and pH=5.75 or so, it should take only a very small amount of additional acid to get to pH=7.