Question

A - B

100. mL   of 0.200 M HCl is titrated with 0.250 M NaOH.

Part A

What is the pH of the solution after 50.0 mL of base has been added?

Express the pH numerically.

pH =

SubmitHintsMy AnswersGive UpReview Part

Part B

What is the pH of the solution at the equivalence point?

Express the pH numerically.

pH =

SubmitHintsMy AnswersGive UpReview Part

Answer 1

A)
This question involves strong acid and strong base
mol of NaOH added = M*V = 0.250 M * 50.0 mL = 12.5 mmol
mol of HCl added = M*V = 0.200 M * 100 mL = 20 mmol

12.5 mmol of each will react
remaining mol of HCl = 20.0 mmol - 12.5 mmol = 7.5 mmol

final volume = 100 mL + 50 mL = 150 mL

[H+] = number of mol / total volume
= 7.5 mmol / 150 mL
= 0.05 M

use:
pH = -log [H+]
= -log (0.05)
= 1.30
Answer: 1.30

B)
At equivalence equal moles of acid and base have been added
pH will be 7
Answer: 7