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 MHBr solution after the following additions of acid: (a) 2.00 ml. pH- (b) 29.00 ml. pH- (c) 35.00 ml

help???steps please

Answer 1

a)when 2.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 2 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

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

mol(HBr) = 0.1 M * 2 mL = 0.2 mmol

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

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

We have:

mol(HBr) = 0.2 mmol

mol(KOH) = 3 mmol

0.2 mmol of both will react

remaining mol of KOH = 2.8 mmol

Total volume = 32.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.8 mmol/32.0 mL

= 8.75*10^-2 M

use:

pOH = -log [OH-]

= -log (8.75*10^-2)

= 1.058

use:

PH = 14 - pOH

= 14 - 1.058

= 12.942

Answer: 12.94

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 35.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 35 mL

M(KOH) = 0.1 M

V(KOH) = 30 mL

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

mol(HBr) = 0.1 M * 35 mL = 3.5 mmol

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

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

We have:

mol(HBr) = 3.5 mmol

mol(KOH) = 3 mmol

3 mmol of both will react

remaining mol of HBr = 0.5 mmol

Total volume = 65.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.5 mmol/65.0 mL

= 7.692*10^-3 M

use:

pH = -log [H+]

= -log (7.692*10^-3)

= 2.1139

Answer: 2.11