Question

A 96.0 mL sample of 0.0300 M HCl is titrated with 0.0600 M KOH solution. Calculate the pH after the following volumes of base have been added.

(a) 18.7 mL

pH =

(b) 45.6 mL

pH =

(c) 48.0 mL

pH =

(d) 48.5 mL

pH =

(e) 85.0 mL

pH =

Answer 1

a)when 18.7 mL of KOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 96 mL

M(KOH) = 0.06 M

V(KOH) = 18.7 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.03 M * 96 mL = 2.88 mmol

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

mol(KOH) = 0.06 M * 18.7 mL = 1.122 mmol

We have:

mol(HCl) = 2.88 mmol

mol(KOH) = 1.122 mmol

1.122 mmol of both will react

remaining mol of HCl = 1.758 mmol

Total volume = 114.7 mL

[H+]= mol of acid remaining / volume

[H+] = 1.758 mmol/114.7 mL

= 1.533*10^-2 M

use:

pH = -log [H+]

= -log (1.533*10^-2)

= 1.8145

Answer: 1.81

b)when 45.6 mL of KOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 96 mL

M(KOH) = 0.06 M

V(KOH) = 45.6 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.03 M * 96 mL = 2.88 mmol

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

mol(KOH) = 0.06 M * 45.6 mL = 2.736 mmol

We have:

mol(HCl) = 2.88 mmol

mol(KOH) = 2.736 mmol

2.736 mmol of both will react

remaining mol of HCl = 0.144 mmol

Total volume = 141.6 mL

[H+]= mol of acid remaining / volume

[H+] = 0.144 mmol/141.6 mL

= 1.017*10^-3 M

use:

pH = -log [H+]

= -log (1.017*10^-3)

= 2.9927

Answer: 2.99

c)when 48.0 mL of KOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 96 mL

M(KOH) = 0.06 M

V(KOH) = 48 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.03 M * 96 mL = 2.88 mmol

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

mol(KOH) = 0.06 M * 48 mL = 2.88 mmol

We have:

mol(HCl) = 2.88 mmol

mol(KOH) = 2.88 mmol

2.88 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

d)when 48.5 mL of KOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 96 mL

M(KOH) = 0.06 M

V(KOH) = 48.5 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.03 M * 96 mL = 2.88 mmol

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

mol(KOH) = 0.06 M * 48.5 mL = 2.91 mmol

We have:

mol(HCl) = 2.88 mmol

mol(KOH) = 2.91 mmol

2.88 mmol of both will react

remaining mol of KOH = 3*10^-2 mmol

Total volume = 144.5 mL

[OH-]= mol of base remaining / volume

[OH-] = 3*10^-2 mmol/144.5 mL

= 2.076*10^-4 M

use:

pOH = -log [OH-]

= -log (2.076*10^-4)

= 3.6827

use:

PH = 14 - pOH

= 14 - 3.6827

= 10.3173

Answer: 10.32

e)when 85.0 mL of KOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 96 mL

M(KOH) = 0.06 M

V(KOH) = 85 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.03 M * 96 mL = 2.88 mmol

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

mol(KOH) = 0.06 M * 85 mL = 5.1 mmol

We have:

mol(HCl) = 2.88 mmol

mol(KOH) = 5.1 mmol

2.88 mmol of both will react

remaining mol of KOH = 2.22 mmol

Total volume = 181.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.22 mmol/181.0 mL

= 1.227*10^-2 M

use:

pOH = -log [OH-]

= -log (1.227*10^-2)

= 1.9113

use:

PH = 14 - pOH

= 14 - 1.9113

= 12.0887

Answer: 12.09