Question

A 74.0 mL sample of 0.0400 M HNO3 is titrated with 0.0800 M LiOH solution. Calculate the pH after the following volumes of base have been added.

(a) 11.8 mL pH =

(b) 36.3 mL pH =

(c) 37.0 mL pH =

(d) 38.9 mL pH =

(e) 59.2 mL pH =

Answer 1

a)when 11.8 mL of LiOH is added

Given:

M(HNO3) = 0.04 M

V(HNO3) = 74 mL

M(LiOH) = 0.08 M

V(LiOH) = 11.8 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.04 M * 74 mL = 2.96 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.08 M * 11.8 mL = 0.944 mmol

We have:

mol(HNO3) = 2.96 mmol

mol(LiOH) = 0.944 mmol

0.944 mmol of both will react

remaining mol of HNO3 = 2.016 mmol

Total volume = 85.8 mL

[H+]= mol of acid remaining / volume

[H+] = 2.016 mmol/85.8 mL

= 2.35*10^-2 M

use:

pH = -log [H+]

= -log (2.35*10^-2)

= 1.629

Answer: 1.63

b)when 36.3 mL of LiOH is added

Given:

M(HNO3) = 0.04 M

V(HNO3) = 74 mL

M(LiOH) = 0.08 M

V(LiOH) = 36.3 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.04 M * 74 mL = 2.96 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.08 M * 36.3 mL = 2.904 mmol

We have:

mol(HNO3) = 2.96 mmol

mol(LiOH) = 2.904 mmol

2.904 mmol of both will react

remaining mol of HNO3 = 5.6*10^-2 mmol

Total volume = 110.3 mL

[H+]= mol of acid remaining / volume

[H+] = 5.6*10^-2 mmol/110.3 mL

= 5.077*10^-4 M

use:

pH = -log [H+]

= -log (5.077*10^-4)

= 3.2944

Answer: 3.29

c)when 37.0 mL of LiOH is added

Given:

M(HNO3) = 0.04 M

V(HNO3) = 74 mL

M(LiOH) = 0.08 M

V(LiOH) = 37 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.04 M * 74 mL = 2.96 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.08 M * 37 mL = 2.96 mmol

We have:

mol(HNO3) = 2.96 mmol

mol(LiOH) = 2.96 mmol

2.96 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

d)when 38.9 mL of LiOH is added

Given:

M(HNO3) = 0.04 M

V(HNO3) = 74 mL

M(LiOH) = 0.08 M

V(LiOH) = 38.9 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.04 M * 74 mL = 2.96 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.08 M * 38.9 mL = 3.112 mmol

We have:

mol(HNO3) = 2.96 mmol

mol(LiOH) = 3.112 mmol

2.96 mmol of both will react

remaining mol of LiOH = 0.152 mmol

Total volume = 112.9 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.152 mmol/112.9 mL

= 1.346*10^-3 M

use:

pOH = -log [OH-]

= -log (1.346*10^-3)

= 2.8709

use:

PH = 14 - pOH

= 14 - 2.8709

= 11.1291

Answer: 11.13

e)when 59.2 mL of LiOH is added

Given:

M(HNO3) = 0.04 M

V(HNO3) = 74 mL

M(LiOH) = 0.08 M

V(LiOH) = 59.2 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.04 M * 74 mL = 2.96 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.08 M * 59.2 mL = 4.736 mmol

We have:

mol(HNO3) = 2.96 mmol

mol(LiOH) = 4.736 mmol

2.96 mmol of both will react

remaining mol of LiOH = 1.776 mmol

Total volume = 133.2 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.776 mmol/133.2 mL

= 1.333*10^-2 M

use:

pOH = -log [OH-]

= -log (1.333*10^-2)

= 1.8751

use:

PH = 14 - pOH

= 14 - 1.8751

= 12.1249

Answer: 12.12