Question

Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid:

38.00ml=

Answer 1

Given:
M(HBr) = 0.1 M
V(HBr) = 38 mL
M(KOH) = 0.1 M
V(KOH) = 30 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 38 mL = 3.8 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 30 mL = 3 mmol


We have:
mol(HBr) = 3.8 mmol
mol(KOH) = 3 mmol
3 mmol of both will react
remaining mol of HBr = 0.8 mmol
Total volume = 68.0 mL

[H+]= mol of acid remaining / volume
[H+] = 0.8 mmol/68.0 mL
= 1.176*10^-2 M


use:
pH = -log [H+]
= -log (1.176*10^-2)
= 1.9294
Answer: 1.93