pH of a buffer solution, how would you find?

Question

I have titration curve Weak acid x Strong Base
Potassium Hydrogen Phthalate x NaOH
acid conc = 0.100 mol dm-3
base conc = 0.101 mol dm-3
The acid was added to the base (40ml) (whole titration)
19.6 ml is the equivalence point
But i do not know how to find the pH of the buffer solution produced
i was told to use this formula :
pH=Pka + lg [A-] / [HA]
The equation is
KHC8H4O4 (aq) + NaOH (aq) ==> KNaC8H4O4 (aq) + H2O(l
how you find the pKA? because i was told you arent able to find one on a pH curve where the acid was added to the alkali, so that it starts from a high pH.

I did take a go at solving this, this is what i managed to get:
ph= pka +log (a-) /(ha)
5.5 = 5.4 + log (0.063) / (0.051)
i took a guess for the pka, i divided the equivalence point by 2 which equals 9.8, and because my graph is the other way round...(this may have been wrong)
i added 9.8 and 19.6, which is 29.4...so at 29.4 ml the pH for that is 5.4, and i assume 5.4 is the pka...=S
am i wrong?

Which answers are wrong? =S

oh i see.....my answers are correct lol ty

Answer 1

If you added acid to base, the titration curve should go downhill, with a steep vertical portion. You noted that this was at 19.6. Add half again, which you have done, and pH = pKa.
So all your work is fine, and the above two answers are wrong. This is an acid buffer, and not a base buffer.

Answer 2

in case you have 10 ml of the buffer, then you certainly've 0.002 moles of lactic acid and nil.003 moles of sodium lactate recent. once you upload 10 ml of 0.01 M HCl, you're including 0.0001 moles of H+, which will react with the lactate to produce lactic acid. you will now have 0.0021 moles lactic acid and nil.0029 moles lactate, yet your quantity has now extra beneficial to 20 mL. nonetheless, because of the fact that they are interior the comparable answer, you ought to use the form of moles interior the Henderson-Hasselbach equation interior the ratio of [base]/[acid] to make certain pH. pH = pKa + log [base]/[acid] = 3.80 5 + log 0.0029/0.0021 = 3.ninety 9 I exceptionally suspect that the pH of the buffer you calculated ahead of the addition of HCl is faulty, or the concentrations are incorrect. If pKa is 3.80 5, your pH can not be decrease than that if the concentration of the backside of the conjugate pair is bigger than the acid.

Answer 3

At the equivalence point, the [A-]=[HA]. So, the pH at the equivalence point = pKa. This comes from the Henderson-Hasselbalch equation that you have. If [A-]=[HA], then that ratio is 1. The log of 1=0. So, pH=pKa.

Hope this helps.

Answer 4

You are dealing with a base so when you figure out the pH of 5.4 you need to take 14-5.4 to figure it out.

8.6 should be your answer. Other than that you're work looks OK.