Question

Be sure to answer all parts.

Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base:

(a) 20.00 mL

pH =

(b) 39.40 mL

pH =

(c) 52.00 mL

pH =

Answer 1

1)when 20.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 20 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

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

mol(NaOH) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

remaining mol of HCl = 2 mmol

Total volume = 60.0 mL

[H+]= mol of acid remaining / volume

[H+] = 2 mmol/60.0 mL

= 3.333*10^-2 M

use:

pH = -log [H+]

= -log (3.333*10^-2)

= 1.4771

2)when 39.4 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 39.4 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

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

mol(NaOH) = 0.1 M * 39.4 mL = 3.94 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 3.94 mmol

3.94 mmol of both will react

remaining mol of HCl = 6*10^-2 mmol

Total volume = 79.4 mL

[H+]= mol of acid remaining / volume

[H+] = 6*10^-2 mmol/79.4 mL

= 7.557*10^-4 M

use:

pH = -log [H+]

= -log (7.557*10^-4)

= 3.1217

3)when 52.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 52 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

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

mol(NaOH) = 0.1 M * 52 mL = 5.2 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 5.2 mmol

4 mmol of both will react

remaining mol of NaOH = 1.2 mmol

Total volume = 92.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.2 mmol/92.0 mL

= 1.304*10^-2 M

use:

pOH = -log [OH-]

= -log (1.304*10^-2)

= 1.8846

use:

PH = 14 - pOH

= 14 - 1.8846

= 12.1154