Question

Consider the titration of a 38.0 mL sample of 0.180 M  HBr with 0.200 M KOH. Determine each of the following:

A.) the initial pH

B.) the volume of added base required to reach the equivalence point

C.) the pH at 12.0 mL of added base

D.) the pH at the equivalence point

E.) the pH after adding 5.0 mL of base beyond the equivalence point

Answer 1

A)when 0.0 mL of KOH is added

Given:

M(HBr) = 0.18 M

V(HBr) = 38 mL

M(KOH) = 0.2 M

V(KOH) = 0 mL

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

mol(HBr) = 0.18 M * 38 mL = 6.84 mmol

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

mol(KOH) = 0.2 M * 0 mL = 0 mmol

We have:

mol(HBr) = 6.84 mmol

mol(KOH) = 0 mmol

0 mmol of both will react

remaining mol of HBr = 6.84 mmol

Total volume = 38.0 mL

[H+]= mol of acid remaining / volume

[H+] = 6.84 mmol/38.0 mL

= 0.18 M

use:

pH = -log [H+]

= -log (0.18)

= 0.7447

Answer: 0.745

B)

Balanced chemical equation is:

HBr + KOH ---> KBr + H2O

Here:

M(HBr)=0.18 M

M(KOH)=0.2 M

V(HBr)=38.0 mL

According to balanced reaction:

1*number of mol of HBr =1*number of mol of KOH

1*M(HBr)*V(HBr) =1*M(KOH)*V(KOH)

1*0.18 M *38.0 mL = 1*0.2M *V(KOH)

V(KOH) = 34.2 mL

Answer: 34.2 mL

C)when 12.0 mL of KOH is added

Given:

M(HBr) = 0.18 M

V(HBr) = 38 mL

M(KOH) = 0.2 M

V(KOH) = 12 mL

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

mol(HBr) = 0.18 M * 38 mL = 6.84 mmol

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

mol(KOH) = 0.2 M * 12 mL = 2.4 mmol

We have:

mol(HBr) = 6.84 mmol

mol(KOH) = 2.4 mmol

2.4 mmol of both will react

remaining mol of HBr = 4.44 mmol

Total volume = 50.0 mL

[H+]= mol of acid remaining / volume

[H+] = 4.44 mmol/50.0 mL

= 8.88*10^-2 M

use:

pH = -log [H+]

= -log (8.88*10^-2)

= 1.0516

Answer: 1.052

D)

This is titration of strong acid and strong base.

At equivalence point, the solution would be neutral and hence pH would be 7

Answer: 7

E)when 39.2 mL of KOH is added

Given:

M(HBr) = 0.18 M

V(HBr) = 38 mL

M(KOH) = 0.2 M

V(KOH) = 39.2 mL

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

mol(HBr) = 0.18 M * 38 mL = 6.84 mmol

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

mol(KOH) = 0.2 M * 39.2 mL = 7.84 mmol

We have:

mol(HBr) = 6.84 mmol

mol(KOH) = 7.84 mmol

6.84 mmol of both will react

remaining mol of KOH = 1 mmol

Total volume = 77.2 mL

[OH-]= mol of base remaining / volume

[OH-] = 1 mmol/77.2 mL

= 1.295*10^-2 M

use:

pOH = -log [OH-]

= -log (1.295*10^-2)

= 1.8876

use:

PH = 14 - pOH

= 14 - 1.8876

= 12.1124

Answer: 12.112