Question

A 250.0 mL solution of 0.255 M C5H6COOH was titrated with 0.750 M NaOH. What is the pH after 105.0 mL of NaOH was added?

Answer 1

Given:

M(C6H5COOH) = 0.255 M

V(C6H5COOH) = 250 mL

M(NaOH) = 0.75 M

V(NaOH) = 105 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.255 M * 250 mL = 63.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.75 M * 105 mL = 78.75 mmol

We have:

mol(C6H5COOH) = 63.75 mmol

mol(NaOH) = 78.75 mmol

63.75 mmol of both will react

excess NaOH remaining = 15 mmol

Volume of Solution = 250 + 105 = 355 mL

[OH-] = 15 mmol/355 mL = 0.0423 M

use:

pOH = -log [OH-]

= -log (4.225*10^-2)

= 1.37

use:

PH = 14 - pOH

= 14 - 1.37

= 12.63

pH = 12.63