Question

# 1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86) is titrated with a...

1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86) is titrated with a 0.100 M NaOH solution. Calculate the pH after the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0, 25.1, 26.0, 28.0, and 30.0 of the NaOH. Plot the results of your calculation, as a pH versus mililiters of NaOH added.

2. Repeat the procedure in problem 1, but the titration of 25.0 mL 0.100 M NH3 (kb= 1.8*10^-5) with 0.100 M HCl.

3.Repeat the procedure in problem 1, but the titration of 25.0 mL 0.100 M pyridine (kb=1.7*10^-9) with 0.100 M hydrofluoric acid (ka= 7.2*10^-4)

4. repeat the procedure in problem 1, but the titration of 25.0 mL 0.100 M NaOH with 0.100 M HNO3.

Q1.

EP is at 25.0mL x0.1M /0.1M = 25.0 mL

V =0.0,

pH = 0.5pKa -0.5logCacid = 1.93 + 0.50 = 2.43

….

V = 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9,

The solution contains the buffer [lactate]/[lactic acid]

pH = pKa + log([lactate]/[lactic acid]) =

= pKa + log (V/(25-V))

Calculate for each V value.

….

V = 25 mL

The solution contains the weak base lactate-. Its concentration is 0.100M/2 = 0.05 M . Its pKb is = 10.14

pOH = 0.5 pKb – 0.5 log 0.05 = 5.07 + 0.65 = 5.72

pH = 8.28

……

V= 25.1, 26.0, 28.0, 30.0mL

The solution contains the excess of a strong base

Vexcess = 0.1   ,   1.0 , 3.0   , 5 mL of NaOH 0.1M diluted in (25 + Vexcess) mL

Calculate for each volume

[HO-] = Vexcess x 0.1 mol/L / (25 + Vexcess)

Then pH = 14 – log[HO-]

…….

Q2

Consider a symetrical situation. Change acid for base and base for acid.

V =0.0,

pOH = 0.5pKb -0.5logCbase = …

pH = 14 – pOH

….

V = 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9,

The solution contains the buffer [NH4+]/[lNH3]

pKa = 14 - pKb

pH = pKa + log([NH3]/[lNH4+]) =

= pKa + log ( (25-V) / V)

Calculate for each V value.

….

V = 25 mL

The solution contains the weak acid NH4+. Its concentration is 0.100M/2 = 0.05 M . Its pKa is = 14 - pKb

pH = 0.5pKa -0.5logCacid = 0.5pKa -0.5log0.05 = …

……

V= 25.1, 26.0, 28.0, 30.0mL

The solution contains the excess of a strong acid

Vexcess = 0.1   ,   1.0 , 3.0   , 5 mL of HCl 0.1M diluted in (25 + Vexcess) mL

Calculate for each volume

[H+] = Vexcess x 0.1 mol/L / (25 + Vexcess)

Then pH = – log[H+]

Q3

V=0

The solution contains the weak base Py. See Q2.

….

V = 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9,

The solution contains the mixture of a weak base Py and a weak acid HF. Ignore Hpy+ and F-.

pH = (pKa1+ pKa2) /2 + log([Py]/[HF]) =

= (pKa1+ pKa2) /2 + log ( (25-V)/V)

Calculate for each V value.

……

V= 25 mL

pH = (pKa1+ pKa2) /2

…………..

V= 25.1, 26.0, 28.0, 30.0mL

Consider the excess of the weak acid HF

pH = 0.5pKa -0.5logCacid

where C = Vexcess x 0.1 mol/L / (25 + Vexcess)

Q4

V=0 mL

[HO-] = 0.1 M   pOH =1        pH=13

V = 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9,

pOH= -log Cbase

where Cbase = 0.1M x (25-V) /(25+V)

Calculate for each V value.

………..

V= 25 mL

[H+] = [HO-]

pH = 7

….

V= 25.1, 26.0, 28.0, 30.0mL

Consider the excess of the strong acid.

Vexcess = 0.1   ,   1.0 , 3.0   , 5 mL of HCl 0.1M diluted in (25 + Vexcess) mL

Calculate for each volume

[H+] = Vexcess x 0.1 mol/L / (25 + Vexcess)

Then pH = – log[H+]