Question

A 98.0 mL sample of 0.0200 M HClO₄ is titrated with 0.0400 M RbOH solution. Calculate the pH after the following volumes of base have been added.

(a) 24.5 mL pH =	(b) 47.0 mL pH =	(c) 49.0 mL pH =
(d) 51.0 mL pH =	(e) 77.4 mL pH =

Answer 1

a)when 24.5 mL of RbO4 is added

Given:

M(HClO4) = 0.02 M

V(HClO4) = 98 mL

M(RbO4) = 0.04 M

V(RbO4) = 24.5 mL

mol(HClO4) = M(HClO4) * V(HClO4)

mol(HClO4) = 0.02 M * 98 mL = 1.96 mmol

mol(RbO4) = M(RbO4) * V(RbO4)

mol(RbO4) = 0.04 M * 24.5 mL = 0.98 mmol

We have:

mol(HClO4) = 1.96 mmol

mol(RbO4) = 0.98 mmol

0.98 mmol of both will react

remaining mol of HClO4 = 0.98 mmol

Total volume = 122.5 mL

[H+]= mol of acid remaining / volume

[H+] = 0.98 mmol/122.5 mL

= 8*10^-3 M

use:

pH = -log [H+]

= -log (8*10^-3)

= 2.0969

Answer: 2.10

b)when 47.0 mL of RbO4 is added

Given:

M(HClO4) = 0.02 M

V(HClO4) = 98 mL

M(RbO4) = 0.04 M

V(RbO4) = 47 mL

mol(HClO4) = M(HClO4) * V(HClO4)

mol(HClO4) = 0.02 M * 98 mL = 1.96 mmol

mol(RbO4) = M(RbO4) * V(RbO4)

mol(RbO4) = 0.04 M * 47 mL = 1.88 mmol

We have:

mol(HClO4) = 1.96 mmol

mol(RbO4) = 1.88 mmol

1.88 mmol of both will react

remaining mol of HClO4 = 8*10^-2 mmol

Total volume = 145.0 mL

[H+]= mol of acid remaining / volume

[H+] = 8*10^-2 mmol/145.0 mL

= 5.517*10^-4 M

use:

pH = -log [H+]

= -log (5.517*10^-4)

= 3.2583

Answer: 3.26

c)when 49.0 mL of RbO4 is added

Given:

M(HClO4) = 0.02 M

V(HClO4) = 98 mL

M(RbO4) = 0.04 M

V(RbO4) = 49 mL

mol(HClO4) = M(HClO4) * V(HClO4)

mol(HClO4) = 0.02 M * 98 mL = 1.96 mmol

mol(RbO4) = M(RbO4) * V(RbO4)

mol(RbO4) = 0.04 M * 49 mL = 1.96 mmol

We have:

mol(HClO4) = 1.96 mmol

mol(RbO4) = 1.96 mmol

1.96 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

d)when 51.0 mL of RbO4 is added

Given:

M(HClO4) = 0.02 M

V(HClO4) = 98 mL

M(RbO4) = 0.04 M

V(RbO4) = 51 mL

mol(HClO4) = M(HClO4) * V(HClO4)

mol(HClO4) = 0.02 M * 98 mL = 1.96 mmol

mol(RbO4) = M(RbO4) * V(RbO4)

mol(RbO4) = 0.04 M * 51 mL = 2.04 mmol

We have:

mol(HClO4) = 1.96 mmol

mol(RbO4) = 2.04 mmol

1.96 mmol of both will react

remaining mol of RbO4 = 8*10^-2 mmol

Total volume = 149.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 8*10^-2 mmol/149.0 mL

= 5.369*10^-4 M

use:

pOH = -log [OH-]

= -log (5.369*10^-4)

= 3.2701

use:

PH = 14 - pOH

= 14 - 3.2701

= 10.7299

Answer: 10.73

e)when 77.4 mL of RbO4 is added

Given:

M(HClO4) = 0.02 M

V(HClO4) = 98 mL

M(RbO4) = 0.04 M

V(RbO4) = 77.4 mL

mol(HClO4) = M(HClO4) * V(HClO4)

mol(HClO4) = 0.02 M * 98 mL = 1.96 mmol

mol(RbO4) = M(RbO4) * V(RbO4)

mol(RbO4) = 0.04 M * 77.4 mL = 3.096 mmol

We have:

mol(HClO4) = 1.96 mmol

mol(RbO4) = 3.096 mmol

1.96 mmol of both will react

remaining mol of RbO4 = 1.136 mmol

Total volume = 175.4 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.136 mmol/175.4 mL

= 6.477*10^-3 M

use:

pOH = -log [OH-]

= -log (6.477*10^-3)

= 2.1887

use:

PH = 14 - pOH

= 14 - 2.1887

= 11.8113

Answer: 11.81