Question

A 25.00 mL sample of 0.310 M NaOH is titrated with 0.750 M HCl at 25 °C.

Calculate the initial pH before any titrant is added.

Calculate the pH of the solution after 5.00 mL of the titrant is added.

Answer 1

1)when 0.0 mL of HCl is added

Given:

M(HCl) = 0.75 M

V(HCl) = 0 mL

M(NaOH) = 0.31 M

V(NaOH) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.75 M * 0 mL = 0 mmol

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

mol(NaOH) = 0.31 M * 25 mL = 7.75 mmol

We have:

mol(HCl) = 0 mmol

mol(NaOH) = 7.75 mmol

0 mmol of both will react

remaining mol of NaOH = 7.75 mmol

Total volume = 25.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 7.75 mmol/25.0 mL

= 0.31 M

use:

pOH = -log [OH-]

= -log (0.31)

= 0.5086

use:

PH = 14 - pOH

= 14 - 0.5086

= 13.4914

Answer: 13.49

2)when 5.0 mL of HCl is added

Given:

M(HCl) = 0.75 M

V(HCl) = 5 mL

M(NaOH) = 0.31 M

V(NaOH) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.75 M * 5 mL = 3.75 mmol

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

mol(NaOH) = 0.31 M * 25 mL = 7.75 mmol

We have:

mol(HCl) = 3.75 mmol

mol(NaOH) = 7.75 mmol

3.75 mmol of both will react

remaining mol of NaOH = 4 mmol

Total volume = 30.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 4 mmol/30.0 mL

= 0.1333 M

use:

pOH = -log [OH-]

= -log (0.1333)

= 0.8751

use:

PH = 14 - pOH

= 14 - 0.8751

= 13.1249

Answer: 13.12