Question

Calculate the pH in the titration of 20 mL of 0.125 M HCL with 0.250 M NaOH solution after adding 9.60 mL and 10.40 mL of NaOH

Answer 1

1)when 9.6 mL of NaOH is added

Given:

M(HCl) = 0.125 M

V(HCl) = 20 mL

M(NaOH) = 0.25 M

V(NaOH) = 9.6 mL

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

mol(HCl) = 0.125 M * 20 mL = 2.5 mmol

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

mol(NaOH) = 0.25 M * 9.6 mL = 2.4 mmol

We have:

mol(HCl) = 2.5 mmol

mol(NaOH) = 2.4 mmol

2.4 mmol of both will react

remaining mol of HCl = 0.1 mmol

Total volume = 29.6 mL

[H+]= mol of acid remaining / volume

[H+] = 0.1 mmol/29.6 mL

= 3.378*10^-3 M

use:

pH = -log [H+]

= -log (3.378*10^-3)

= 2.4713

Answer: 2.47

2)when 10.4 mL of NaOH is added

Given:

M(HCl) = 0.125 M

V(HCl) = 20 mL

M(NaOH) = 0.25 M

V(NaOH) = 10.4 mL

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

mol(HCl) = 0.125 M * 20 mL = 2.5 mmol

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

mol(NaOH) = 0.25 M * 10.4 mL = 2.6 mmol

We have:

mol(HCl) = 2.5 mmol

mol(NaOH) = 2.6 mmol

2.5 mmol of both will react

remaining mol of NaOH = 0.1 mmol

Total volume = 30.4 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.1 mmol/30.4 mL

= 3.289*10^-3 M

use:

pOH = -log [OH-]

= -log (3.289*10^-3)

= 2.4829

use:

PH = 14 - pOH

= 14 - 2.4829

= 11.5171

Answer: 11.52