Question

37. A 87.0 mL sample of 0.0400 M HBrO₄ is titrated with 0.0800 M NaOH solution. Calculate the pH after the following volumes of base have been added.

Answer 1

37)

a)when 20.9 mL of NaOH is added

Given:

M(HBrO4) = 0.04 M

V(HBrO4) = 87 mL

M(NaOH) = 0.08 M

V(NaOH) = 20.9 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.04 M * 87 mL = 3.48 mmol

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

mol(NaOH) = 0.08 M * 20.9 mL = 1.672 mmol

We have:

mol(HBrO4) = 3.48 mmol

mol(NaOH) = 1.672 mmol

1.672 mmol of both will react

remaining mol of HBrO4 = 1.808 mmol

Total volume = 107.9 mL

[H+]= mol of acid remaining / volume

[H+] = 1.808 mmol/107.9 mL

= 1.676*10^-2 M

use:

pH = -log [H+]

= -log (1.676*10^-2)

= 1.7758

Answer: 1.78

b)when 42.2 mL of NaOH is added

Given:

M(HBrO4) = 0.04 M

V(HBrO4) = 87 mL

M(NaOH) = 0.08 M

V(NaOH) = 42.2 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.04 M * 87 mL = 3.48 mmol

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

mol(NaOH) = 0.08 M * 42.2 mL = 3.376 mmol

We have:

mol(HBrO4) = 3.48 mmol

mol(NaOH) = 3.376 mmol

3.376 mmol of both will react

remaining mol of HBrO4 = 0.104 mmol

Total volume = 129.2 mL

[H+]= mol of acid remaining / volume

[H+] = 0.104 mmol/129.2 mL

= 8.05*10^-4 M

use:

pH = -log [H+]

= -log (8.05*10^-4)

= 3.0942

Answer: 3.09

c)when 43.5 mL of NaOH is added

Given:

M(HBrO4) = 0.04 M

V(HBrO4) = 87 mL

M(NaOH) = 0.08 M

V(NaOH) = 43.5 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.04 M * 87 mL = 3.48 mmol

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

mol(NaOH) = 0.08 M * 43.5 mL = 3.48 mmol

We have:

mol(HBrO4) = 3.48 mmol

mol(NaOH) = 3.48 mmol

3.48 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

d)when 45.2 mL of NaOH is added

Given:

M(HBrO4) = 0.04 M

V(HBrO4) = 87 mL

M(NaOH) = 0.08 M

V(NaOH) = 45.2 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.04 M * 87 mL = 3.48 mmol

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

mol(NaOH) = 0.08 M * 45.2 mL = 3.616 mmol

We have:

mol(HBrO4) = 3.48 mmol

mol(NaOH) = 3.616 mmol

3.48 mmol of both will react

remaining mol of NaOH = 0.136 mmol

Total volume = 132.2 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.136 mmol/132.2 mL

= 1.029*10^-3 M

use:

pOH = -log [OH-]

= -log (1.029*10^-3)

= 2.9877

use:

PH = 14 - pOH

= 14 - 2.9877

= 11.0123

Answer: 11.01

e)when 81.8 mL of NaOH is added

Given:

M(HBrO4) = 0.04 M

V(HBrO4) = 87 mL

M(NaOH) = 0.08 M

V(NaOH) = 81.8 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.04 M * 87 mL = 3.48 mmol

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

mol(NaOH) = 0.08 M * 81.8 mL = 6.544 mmol

We have:

mol(HBrO4) = 3.48 mmol

mol(NaOH) = 6.544 mmol

3.48 mmol of both will react

remaining mol of NaOH = 3.064 mmol

Total volume = 168.8 mL

[OH-]= mol of base remaining / volume

[OH-] = 3.064 mmol/168.8 mL

= 1.815*10^-2 M

use:

pOH = -log [OH-]

= -log (1.815*10^-2)

= 1.7411

use:

PH = 14 - pOH

= 14 - 1.7411

= 12.2589

Answer: 12.26

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