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: (а) 25.00 mL PН- (b) 39.00 mL pH (с) 55.00 mL pH

Answer 1

a)when 25.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 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 * 25 mL = 2.5 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

remaining mol of HCl = 1.5 mmol

Total volume = 65.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1.5 mmol/65.0 mL

= 2.308*10^-2 M

use:

pH = -log [H+]

= -log (2.308*10^-2)

= 1.6368

Answer: 1.64

b)when 39.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 39 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 mL = 3.9 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 3.9 mmol

3.9 mmol of both will react

remaining mol of HCl = 0.1 mmol

Total volume = 79.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.1 mmol/79.0 mL

= 1.266*10^-3 M

use:

pH = -log [H+]

= -log (1.266*10^-3)

= 2.8976

Answer: 2.90

c)when 55.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 55 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 * 55 mL = 5.5 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 5.5 mmol

4 mmol of both will react

remaining mol of NaOH = 1.5 mmol

Total volume = 95.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.5 mmol/95.0 mL

= 1.579*10^-2 M

use:

pOH = -log [OH-]

= -log (1.579*10^-2)

= 1.8016

use:

PH = 14 - pOH

= 14 - 1.8016

= 12.1984

Answer: 12.20