Finding HCl and HOAc concentrations from NaOH titration equivalence points

[Rissj3]

Hi, I'm in lab in which we have to titrate 0.09M NaOH into 25mL of 0.1M HCl/HOAc mixture.

Homework Statement

I need help on how to calculate the concentration of HCl and HOAc in the mixture.

0.09M NaOH
1st equivalence point = 14.3mL
2nd equivalence point = 26.9mL

Difference of two equivalence points, 14.3-26.9= 12.6mL

Homework Equations

I'm thinking to solve this problem I would have to use the dilution formula, M1V1=M2V2

The Attempt at a Solution

M (0.025L) = (0.09M)(0.0143L)
M = 0.05148M HCl

M (0.025L) = (0.09M)(0.0126L)
M = 0.04536M HOAc

Is this correct or am I completely off?

Any help would be appreciated.

Thanks

 

[symbolipoint]

To give a simplified explanation:
The weak acid will be picked at titrant addition to about pH of 4 to 4.5, and both the weak and strong acid will be neutralized at about pH 8 to 9. Your first equivalence point will give you moles of HOAc; your second equivalence point will give you moles of both HOAc and HCl combined. The difference will be for the moles of HCl alone.

 

[Borek]

I'm thinking to solve this problem I would have to use the dilution formula, M1V1=M2V2

Since when titration is identical to dilution?

http://www.titrations.info/titration-calculation

Formula you listed happens to be identical to the one needed in SOME cases. The longer you will believe it works always, the worse for you.