[ItalianChick0188]

So in lab we measured 2 samples of an unknown substance and titrated it  twice with  0.1007 M NaOH using a pH meter model 420A and an Orion pH meter model 310*.

My weight of the unknown for sample 1 was 0.0998 grams. We are asked to calculate the moles of NaOH reacted at the equivalence point.

I made my graphs and found that the equivalence point for Trial 1 was 5.38 mL NaOH with a pH of 8.0.

Would it be 0.0998 g unknown /0.1007 M NaOH?

[Borek]

Reaction equation, please.

[ItalianChick0188]

HA + NaOH ------ A- + H20 + Na+

[ItalianChick0188]

Wait it would be 0.1007 M NaOH * 0.00538 L = 5.42 X 10 ^ -4 moles NaOH.

[ItalianChick0188]

Alright so I calculated the moles of acid reacted and NaOH which was 5.42 X 10 ^ -4 and I also calculated the MW of the acid and I got 184.13 g/M.

I am now asked to calculate (H30+ ) at 50% titration.

My volume of NaOH at 50% titration was calculated to be 5.00 mL and  the pH at 50% was 5.21.


How could I calculate H30+ at 50% titration?

[Borek]

What is pH definition?