Question

Be sure to answer all parts. Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base:

(a) 17.00 mL pH =

(b) 39.20 mL pH =

(c) 48.00 mL pH =

Answer 1

(a) 17.00 mL pH = 1.39

(b) 39.20 mL pH = 3.00

(c) 48.00 mL pH = 11.96

Explanation

(a) moles HCl = (molarity HCl) * (volume HCl)

moles HCl = (0.1000 M) * (40.00 mL)

moles HCl = 4.000 mmol

moles NaOH = (molarity NaOH) * (volume NaOH)

moles NaOH = (0.1000 M) * (17.00 mL)

moles NaOH = 1.700 mmol

moles HCl neutralized = moles NaOH

moles HCl neutralized = 1.700 mol

moles HCl remaining = initial moles HCl - moles HCl neutralized

moles HCl remaining = 4.000 mmol - 1.7 mmol

moles HCl remaining = 2.300 mmol

molarity HCl = (moles HCl remaining) / (total volume)

molarity HCl = (2.300 mmol) / (40.00 mL + 17.00 mL)

molarity HCl = 0.0404 M

pH = -log[H⁺]

pH = -log(0.0404 M)

pH = 1.39