 | Sample Lab Questions Investigation #22:
"Determination of an Ionization Constant"
|
Click HERE for an overview of Investigation 22, compliments of Mr. Alex Redd.
These questions have been taken from lab quizzes given in previous years.
Some of the answers are listed at the end.
Note: The subscripts and superscripts may not appear as such. It depends upon your web browser.
- 1.
- (4pts) Draw the titration curve for a weak acid titrated by a strong base. Label the axes, the half-neutralization point. Graphically indicate the value of pKa.
- 2.
- (1pt) Why was the distilled water boiled before being used to prepare solutions for Inv #24, "Determination of an Ionization Constant"?
- 3.
- You are given a pH meter, a standard buffer solution of pH 4.01, distilled water, solution A (0.001M HCl), and solution B (0.05M acetic acid).
- a. (2pts) Describe briefly the procedure you would follow to measure the pH’s of solutions A and B.
- b. (1pt) What would be the pH of solution A? (Show your calculations).
- c. (2pts) If the pH of solution B is measured to be 2.98, calculate the Ka for acetic acid.
- 4.
- a. (2pts) In Investigation #24, what compound do we use to standardize the NaOH solution (give name and formula)?
- b. (1pt) Why do we need to do this standardization of NaOH solution?
- 5.
- (2pts) 0.5774g of KHP (C8H5O4K)was dissolved in 50 mL of water and 4 drops of phenolphthalein was added. The solution was then titrated with 25.68 mL of a NaOH solution. What is the concentration of the NaOH solution?
- 6.
- A monoprotic acid, HA, dissociates into H+ and A-.
- a. (1pt) If 0.3628 g of HA requires 34.86 mL of 0.1056 M NaOH for titration, what is the molecular mass of HA?
- b. (3pts) If 0.3826 g of HA was dissolved and diluted to make 100.00 mL of aqueous solution and the pH of this solution was 3.56, what is the Ka for HA?
Answers
5. | | 0.1101 M NaOH |
6. | a. | 98.55 g HA/mol |
| b. | [H+] = [A-] = 2.8 x 10-4 M [HA} = 0.03882 M Ka = 2.0 x 10-6 |

To report any corrections, please e-mail Dr. Wendy Keeney-Kennicutt.