Question

A 91.0 mL sample of 0.0300 M HCl is titrated with 0.0600 M ROOH solution. Calculate the pH after the following volumes of base have been added. (a) 16.8 ml (b) 43.2 mL (C) 45.5 mL pH = pH = pH = (d) 46.0 mL () 82.8 ml pH = pH =

Answer 1

1)when 16.8 mL of RbOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 91 mL

M(RbOH) = 0.06 M

V(RbOH) = 16.8 mL

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

mol(HCl) = 0.03 M * 91 mL = 2.73 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 16.8 mL = 1.008 mmol

We have:

mol(HCl) = 2.73 mmol

mol(RbOH) = 1.008 mmol

1.008 mmol of both will react

remaining mol of HCl = 1.722 mmol

Total volume = 107.8 mL

[H+]= mol of acid remaining / volume

[H+] = 1.722 mmol/107.8 mL

= 1.597*10^-2 M

use:

pH = -log [H+]

= -log (1.597*10^-2)

= 1.7966

Answer: 1.80

2)when 43.2 mL of RbOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 91 mL

M(RbOH) = 0.06 M

V(RbOH) = 43.2 mL

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

mol(HCl) = 0.03 M * 91 mL = 2.73 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 43.2 mL = 2.592 mmol

We have:

mol(HCl) = 2.73 mmol

mol(RbOH) = 2.592 mmol

2.592 mmol of both will react

remaining mol of HCl = 0.138 mmol

Total volume = 134.2 mL

[H+]= mol of acid remaining / volume

[H+] = 0.138 mmol/134.2 mL

= 1.028*10^-3 M

use:

pH = -log [H+]

= -log (1.028*10^-3)

= 2.9879

Answer: 2.99

3)when 45.5 mL of RbOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 91 mL

M(RbOH) = 0.06 M

V(RbOH) = 45.5 mL

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

mol(HCl) = 0.03 M * 91 mL = 2.73 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 45.5 mL = 2.73 mmol

We have:

mol(HCl) = 2.73 mmol

mol(RbOH) = 2.73 mmol

2.73 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

4)when 46.0 mL of RbOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 91 mL

M(RbOH) = 0.06 M

V(RbOH) = 46 mL

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

mol(HCl) = 0.03 M * 91 mL = 2.73 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 46 mL = 2.76 mmol

We have:

mol(HCl) = 2.73 mmol

mol(RbOH) = 2.76 mmol

2.73 mmol of both will react

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

Total volume = 137.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 3*10^-2 mmol/137.0 mL

= 2.19*10^-4 M

use:

pOH = -log [OH-]

= -log (2.19*10^-4)

= 3.6596

use:

PH = 14 - pOH

= 14 - 3.6596

= 10.3404

Answer: 10.34

5)when 82.8 mL of RbOH is added

Given:

M(HCl) = 0.03 M

V(HCl) = 91 mL

M(RbOH) = 0.06 M

V(RbOH) = 82.8 mL

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

mol(HCl) = 0.03 M * 91 mL = 2.73 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 82.8 mL = 4.968 mmol

We have:

mol(HCl) = 2.73 mmol

mol(RbOH) = 4.968 mmol

2.73 mmol of both will react

remaining mol of RbOH = 2.238 mmol

Total volume = 173.8 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.238 mmol/173.8 mL

= 1.288*10^-2 M

use:

pOH = -log [OH-]

= -log (1.288*10^-2)

= 1.8902

use:

PH = 14 - pOH

= 14 - 1.8902

= 12.1098

Answer: 12.11