Question

In: Chemistry

Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M...

Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 5.00 mL (b) 29.50 mL (c) 36.00 mL

Solutions

Expert Solution

a)

Given:

M(HBr) = 0.1 M

V(HBr) = 5 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 5 mL = 0.5 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 0.5 mmol

mol(KOH) = 3 mmol

0.5 mmol of both will react

remaining mol of KOH = 2.5 mmol

Total volume = 35.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.5 mmol/35.0 mL

= 0.0714 M

use:

pOH = -log [OH-]

= -log (7.143*10^-2)

= 1.1461

use:

PH = 14 - pOH

= 14 - 1.1461

= 12.8539

b)

Given:

M(HBr) = 0.1 M

V(HBr) = 29.5 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 29.5 mL = 2.95 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 2.95 mmol

mol(KOH) = 3 mmol

2.95 mmol of both will react

remaining mol of KOH = 0.05 mmol

Total volume = 59.5 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.05 mmol/59.5 mL

= 0.0008 M

use:

pOH = -log [OH-]

= -log (8.403*10^-4)

= 3.0755

use:

PH = 14 - pOH

= 14 - 3.0755

= 10.9245

c)

Given:

M(HBr) = 0.1 M

V(HBr) = 36 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 36 mL = 3.6 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HBr) = 3.6 mmol

mol(KOH) = 3 mmol

3 mmol of both will react

remaining mol of HBr = 0.6 mmol

Total volume = 66.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.6 mmol/66.0 mL

= 0.0091 M

use:

pH = -log [H+]

= -log (9.091*10^-3)

= 2.0414


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