Question

Be sure to answer all parts. Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after the following additions of base: (a) 20.00 mL pH = (b) 39.10 mL pH = (c) 51.00 mL pH=

steps needed with answers please

Answer 1

a)when 20.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 20 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react

remaining mol of HCl = 2 mmol

Total volume = 60.0 mL

[H+]= mol of acid remaining / volume

[H+] = 2 mmol/60.0 mL

= 3.333*10^-2 M

use:

pH = -log [H+]

= -log (3.333*10^-2)

= 1.4771

Answer: 1.48

b)when 39.1 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 39.1 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 39.1 mL = 3.91 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 3.91 mmol

3.91 mmol of both will react

remaining mol of HCl = 9*10^-2 mmol

Total volume = 79.1 mL

[H+]= mol of acid remaining / volume

[H+] = 9*10^-2 mmol/79.1 mL

= 1.138*10^-3 M

use:

pH = -log [H+]

= -log (1.138*10^-3)

= 2.9439

Answer: 2.94

c)when 51.0 mL of NaOH is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40 mL

M(NaOH) = 0.1 M

V(NaOH) = 51 mL

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

mol(HCl) = 0.1 M * 40 mL = 4 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 51 mL = 5.1 mmol

We have:

mol(HCl) = 4 mmol

mol(NaOH) = 5.1 mmol

4 mmol of both will react

remaining mol of NaOH = 1.1 mmol

Total volume = 91.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.1 mmol/91.0 mL

= 1.209*10^-2 M

use:

pOH = -log [OH-]

= -log (1.209*10^-2)

= 1.9176

use:

PH = 14 - pOH

= 14 - 1.9176

= 12.0824

Answer: 12.08