Question

Be sure to answer all parts. 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) 4.00 mL pH = (b) 29.00 mL pH = (c) 36.00 mL pH =

Answer 1

a)when 4.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 4 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

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

mol(HBr) = 0.1 M * 4 mL = 0.4 mmol

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

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

We have:

mol(HBr) = 0.4 mmol

mol(KOH) = 3 mmol

0.4 mmol of both will react

remaining mol of KOH = 2.6 mmol

Total volume = 34.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.6 mmol/34.0 mL

= 7.647*10^-2 M

use:

pOH = -log [OH-]

= -log (7.647*10^-2)

= 1.1165

use:

PH = 14 - pOH

= 14 - 1.1165

= 12.8835

Answer: 12.88

b)when 29.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 29 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

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

mol(HBr) = 0.1 M * 29 mL = 2.9 mmol

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

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

We have:

mol(HBr) = 2.9 mmol

mol(KOH) = 3 mmol

2.9 mmol of both will react

remaining mol of KOH = 0.1 mmol

Total volume = 59.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.1 mmol/59.0 mL

= 1.695*10^-3 M

use:

pOH = -log [OH-]

= -log (1.695*10^-3)

= 2.7709

use:

PH = 14 - pOH

= 14 - 2.7709

= 11.2291

Answer: 11.23

c)when 36.0 mL of HBr is added

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

= 9.091*10^-3 M

use:

pH = -log [H+]

= -log (9.091*10^-3)

= 2.0414

Answer: 2.04