InClass03 AcidBase 105F96

Chem 105 InClass Work

Acids and Bases

October 17, 1996 Name ___Zellmer-Key_____

Here is a table of Acid Dissociation Constants at 25deg. C. Values are given in the "E" style of scientific notation, i.e. 1.75E-05 is 1.75 x 10-5.

Name

K1 (or Ka)
K2
K3
K4
Acetic acid
1.75E-05



Ammonium ion
5.70E-10



Arsenic acid (H3AsO4)
5.8E-03
1.10E-07
3.2E-12

Benzoic acid
6.8E-05



Pyridinium ion (C5H5NH+)
5.9E-06



Carbonic acid
4.45E-07
4.69E-11


o-Phthalic acid
1.12E-03
3.90E-06


1. For each substance below, state if it is a Strong Acid (SA), Strong Base (SB), Weak Acid (WA), Weak Base (WB), Buffer (Buffer), or Amphiprotic (Amphiprotic). As a first approximation, certain species of polyprotic acids and some mixtures can be considered to fall into one of these categories also.

Substance

What is it?
Substance
What is it?
Acetic Acid
WA
CH3COONa
WB
"KHP" (P=phthalate)
Amphiprot
Pyridine hydrochloride
WA
CO2 in water
WA
HNO3
SA
Sodium Benzoate
WB
0.001M KOH
SB
0.1M NH3 + 0.2M NH4+
Buffer
0.1M NaOH + 0.01M NH3
~SB

2. Calculate the pH of 0.001 M KOH.

Strong Base problem

[OH-] = Csb = 1.0 x 10-3; pOH=3.0; pH = 14 - 3.0 = 11.0

3. Calculate the pH of 0.15M sodium acetate.

Weak Base problem (salt of acetic acid)

Kb = 10-14/1.75 x 10-5 = 5.71 x 10-10

[OH-] = sqrt(5.71E-10 *(0.15M - [OH-])) = 9.26E-06

This is close enough.

pOH = 5.03; pH = 14 - 5.03 = 8.96

4. Calculate the pH of 0.001M Benzoic Acid.

This is a Weak Acid problem.

[H+]1 = sqrt(6.8E-05*(0.001 - [H+] as 0)) = 2.61E-04

[H+]2 = sqrt(6.8E-05*(0.001 - 2.61E-04)) = 2.24E-04

[H+]3 = sqrt(6.8E-05*(0.001 - 2.24E-04)) = 2.30E-04

[H+]4 = sqrt(6.8E-05*(0.001 - 2.30E-04)) = 2.29E-04

pH = -log(2.29 x 10-4) = 3.64

5. 50.0 mL of a buffer consisting of 0.100M acetic acid and 0.200 M sodium acetate is about to be hit with 2.0 mL of 0.100 M HCl.

A. Calculate the pH of the buffer before the HCl is added.

pH = 4.76 + log(0.200M/0.100M) = 5.06

B. Use an Initial/Final Spreadsheet to calculate the pH of the solution after the HCl has been added. (It may or may not still be a buffer.)

Balanced Chemical Reaction: CH3COO- + HCl = CH3COOH + Cl-

CH3COOH

CH3COO-
H+ (HCl)
Volume
Initial mmoles

5.00 mmole

10.00 mmole
0.2 mmole
50+2 mL
Final mmoles

5.2 mmole

9.80 mmole
0 mmole
52 mL
Molarity

5.2 mmole/52mL

9.8mmole/52mL
0 M

pH = 4.76 + log(9.8mmole/5.2mmole) = 5.04

Note that the pH was changed very little. The buffer has done its job.

6. Sketch the titration curve of 50.0 mL of 0.0500 M pyridinium chloride with 0.100 M NaOH.

Ka = 5.90 x 10-6; pKa = 5.23

Calculate mL to equivalence point:

0.05 mmole pyr

50 mL
1 NaOH
mL
mL

1 pyr
0.1 mmole NaOH

= 25.0 mL

The buffer region will center on pH = pKa = 5.23.

The final pH will be that of diluted 0.1 M NaOH, or about 12.

Since the break is 12 - 5.2 = 6.8 which is greater than 4, it will be sharp.