Calculate the pH during a titration when 10.00mL of 0.098 M HCl has reacted with 9.35 mL of 0.02 M NaOH?

1 Answer
Mar 31, 2018

pH = 1.388

Explanation:

By definition

pH = -log[HCl]
So,

INITIALLY
pH = -log[.098] = 1.009

Neutralization equation:

HCl + NaOH "----->" NaCL + H_2 O

color(blue)"Purpose": to calculate the concentration of [HCl] color(red)"left" after the addition of 9.35 mL of 0.02 M NaOH.

Moles of HCl:
(.098 "mol"/L) .010 L = .00098 mol of HCl

Moles of NaOH added
(.02 "mol"/L) .00938 L = .0001876 mol of NaOH

Stoichiometry molar report is HCl = NaOH

Moles of left of HCl = .00098 - .0001878 = .0007924"

New concentration of HCl = "moles left" / "total volume created"

HCl = (.0007924 "_mol") / "(.010 + .00938) L" = .04095 M

FINALLY

pH = -log[.04095] = 1.388