Question

I need help with part c and d please. I would like to see steps to understand.

Equal volumes of two acids are titrated with 0.10M NaOH resulting in the two titration curves shown in the following figure(Figure 1) Figure < 1 of 1 > PH 0 10 20 30 ml NaOH 40 50

Answer 1

12 10 half equivalence point 6 4 30 40 50 10 20 0 mL NaOH 2 0 Hа

at half equivalence point pH = pKa

part C)

equivalence point volume = 25 mL and half equivalence point 12.5 mL

here pH = pKa

pKa = 7

part D)

at equivalence point : moles of acid = moles of base

moles of base = 35 x 0.1 /1000 = 3.5 x 10^-3

moles of acid = 3.5 x 10^-3

mass / molar mass = 3.5 x 10^-3

0.75 / molar mass = 3.5 x 10^-3

molar mass = 214 g / mol