Equilibrium constant, Ka

1
Equilibrium constant, Ka

• By Soo Jeon

2
background

• Most weak acids dissociate 5 or less, that is,
  95 or more of the acid remains as HA. The
  smaller the value of Ka, the weaker the acid.

3
Background ..

• Monoprotic acids are those acids that are able to
  donate one proton per molecule during the process
  of dissociation (sometimes called ionization) as
  shown below (symbolized by HA)
• HA(aq)    H2O(l)  -gt     H3O(aq)    A-(aq)

4
scenario

• A student prepared three solutions by adding
  12.00, 18.5, and 35.00ml of 7.5x10-2M NaOH
  solution to 50.00ml of a 0.100M solution of a
  weak monoprotic acid, HA. The solutions were
  labeled X,Y,and Z respectively. Each of the
  solutions were diluted to a total volume of
  100.00ml with distilled water. The pH readings of
  these solutions, obtained through the use of a
  calibrated pH meter, were X,Y, Z. We are going to
  determine Ka for a monoprotic weak acid.

5
Purpose

• This experiment will determine Ka for a
  monoprotic weak acid using pH value and A-ion
  concentration.

6
Materials

• 7.5x10-2M NaOH solution
• 0.100M solution of a weak monoprotic acid
• distilled water
• calibrated pH meter
• 100 ml of Three beakers

7
pH of each solution

• Using the calibrated pH meter, find pH of each
  solution.
• X 6.50
• Y 6.70
• Z 7.10

8
Calculations

• Convert pH to equivalent H30 concentration.
• 1) pH -log H
• H30 10-pH
• X 10-6.50 3.16 x 10-7M H30
• Y 10-6.70 2.00 x 10-7M H30
• Z 10-7.10 7.94 x 10-8M H30

9
2. Calculate the number of moles of acid added

• The moles of acid was constant for each solution.
• 50.00ml x 1L x 0.100 moles HA
• 1 1000ml 1 L
• 5.00 x 10-3 mol HA

10
3. Calculate the number of moles of OH- added

• X 12.00ml x 1L x 7.50 x 10-2 mol OH-
• 1 1000ml 1 L
• 9.00 x 10-4 mol OH-
• Y 18.50ml x 1L x 7.50 x 10-2mol OH-
• 1 1000ml 1L
• 1.39 x 10-3 mol OH-
• Z 35.00ml x 1L x 7.50 x 10-2mol OH-
• 1 1000ml 1L
• 2.69 x 10-3 mol OH-

11
4. Determine the concentration of HA and A- ion

• X HA initial moles HA moles OH- added
• volume of
  solution
• 5.00 x 10-3 - 9.00 x 10-4
• 0.100 L
• 4.10 x 10-2 Molarity
• Moles A- formed moles OH- added
• A- 9.00 x 10-4 mol A- 9.00 x 10-3
  Molarity
• 0.100L

12

• Y HA 5.00 x 10-3 1.39 x 10-3
• 0.1000L
• 3.61 x 10-2 Molarity
• A- 1.39 x 10-3 mol A-
• 0.1000L
• 1.39 x 10-2 Molarity

13

• Z HA 5.00 x 10-3 - 2.63 x 10-3
• 0.10000L
• 2.37 x 10-2 Molarity
• A- 2.63 x 10-3 mol A-
• 0.10000L
• 2.63 x 10-2 Molarity

14
5. Reciprocal of A- ion concentration

• To show that the concentration is increasing, we
  need to flip the A- ion concentration.
• X 1/(9.00 x 10-3) 1.11 x 10-2 M-1
• Y 1/(1.39 x 10-2) 7.19 x 101 M-1
• Z 1/(2.63 x 10-2) 3.80 x 101 M-1

15
6. Reciprocal of A- ion concentration against
H3O ion concentration

• If it is not reciprocal of A- ion concentration,
• then the slope (Ka) would be negative value.

16
7. Determine the slope of the line

• Slope ?Y ?1/A-
• ?X ?H30
• 1.11 x 102 3.80 x 101 3.08 x 108
• (3.16 x 10-7) (7.94 x 10-8)

17
8. Determine the initial concentration of the
acid, HA.

• 5.00 x 10-3 mol HA 5.00 x 10-2 M
• 0.10000L

18
9. Determine the Ka for the weak acid.

• 1 ?H30 (H30) (A-)
• Slope ?1/A-
• Ka 1 1___________
  
  
• HA x slope (5.00 x 10-2)(3.08 x
  101)
• 6.49 x 10-8 M

19
Conclusion

• In this lab, I measured pH values of each
  solution and then determined the concentrations
  of HA and A- ion. I calculated the slope of the
  reciprocal of A- ion concentration on the
  ordinate against the H3O ion concentration and
  got the Ka for the weak acid, 6.49 x 10-8 M.

20
Citation

• http//answers.yahoo.com/question/index?qid200902
  01045717AAjsY1g
• http//wb7.itrademarket.com/pdimage/15/400615_cybe
  rscanpc300.jpg

21
Image citations

• Slide 6 http//www.micglobal.co.uk/images/beake
  rs_14000BGYR.jpg
• Slide 7 http//wb6.itrademarket.com/pdimage/15
  /400615_cyberscanpc300.jpg
• Slide 8 http//www.flickr.com/photos/inthe_aero
  plane/526874125/