Question

A 25.00 mL sample of 0.280 M NaOH analyte was titrated with 0.750 M HCl at 25 °C.

Calculate the initial pH before any titrant was added.

​Calculate pH after 5.0 ml of titrant was added

Answer 1

Initially:pH = 13.45

pH after 5.0 ml of titrant = 13.08

THE SOLUTIONS:

During the titration of a strong base with a strong acid:

The initial pH of the solution, that is, without adding the acid titrant, is mainly due to the concentration of the analyte.

Hence,

Volume of 0.750 M HCl = 0,

then [OH-] = [NaOH]

[OH-] = 0.280 M

pOH = -log[OH-] = -log (0.28) = 0.553

pH = 14 –pOH = 14 - 0.553 = 13.447 =13.45

(a) Volume of 0.750 M HCl = 5.0 mL,

so,
5.0 mL * 0.750 M HCl = 3.75 mmol HCl

25.0 mL * 0.280 M NaOH = 7 mmol NaOH

the difference is 3.5 mmol NaOH (in excess)

To solve for the concentration, we must find the TOTAL VOLUME of the solution:

TOTAL VOLUME = 25.0 mL + 5.0 mL = 30.0 mL

[OH-] = 3.5 mmol NaOH / 30.0 mL = 0.12

pOH = -log[OH-] = log (0.12) = 0.92

pH = 14 – pOH = 14 – 0.92 = 13.08