calculation for each sample CHEMICAL EQUILIBRIUM II (Lab #3) (continued) pH OF S

Question

calculation for each sample

CHEMICAL EQUILIBRIUM II (Lab #3) (continued) pH OF SOLUTIONS and IONIZATION CONSTANTS OF WEAK ACIDS AND BASES PROCEDURE 1. Turn on the pH meter and let it warm up for at least fifteen (15) minutes. 2. Connect the electrode probes to the pH meter, and dip plH probe bulb in the bufer solution. 3. When the instrument is warm and ready, standardize it. Switch to the "Read" mode and adjust the meter to indicate the pH value of the buffer using the "Calibrate" dial. Your instructor will show you how! 4. After the pH meter has been standardized, the instrument is ready to measure the pH of your solutions. You must switch the meter to standby before the electrodes are removed from the solution. The electrodes must be rinsed with water when changing and measuring different solutions. 5. Use the solution containing 0.1MAcetic Acid. The concentration (molarity) of the acid is written on the reagent bottle and will be the initial concentration of the acid. for each solution. Discuss and compare the Ka to that value for acetic acid in the textbook. Solution 1: a sample of the original 0.1 M Acetic Acid 6. Prepare the following solutions, and measure and record their pH. Then calculate the Ka of the weak acid 2.65 7. Solution 2: Dilute the 0.1 M Acetic Acid 1:10 by adding 10 mL of Acetic Acid to 90 mL water Calculate the new concentration. .20 Solution 3: Add 20 mL of0.1 M Acetic Acid to 20 mL of 0.1 M Sodium Acetate (the sodium salt of the weak acid), Calculate the new concentrations of the acetic acid and the salt of the acid Solution 4: Add 5mL of the 0.1 M Acetic Acid to 25 mL of the 0.1 M Sodium Acetate plus 20 mL of U.54 water. Calculate the concentration of the acid and the salt of the acid. 5.25

Explanation / Answer

Sample 1.

[Acetic acid] = 0.1 M and pKa of acetic acid = 4.7

pH = 1/2 (pKa - Log[Acetic acid])

= 1/2 * {4.7 - Log(0.1)}

= 1/2 * {4.7-(-1)}

= 1/2 * (5.7)

= 2.85

Sample 2.

[Acetic acid] = 0.1*10/100 = 0.01 M

Now, pH = 1/2 * {4.7 - Log(0.01)}

= 1/2 * {4.7-(-2)}

= 1/2 * (6.7)

= 3.35

Sample 3.

According to Henderson-Hasselbulch equation:

[Acetic acid] = 0.1*20/40 = 0.05 M

[sodium acetate] = 0.1*20/40 = 0.05

pH = pKa + Log([sodium acetate]/[acetic acid])

= 4.7 + Log(0.05/0.05)

= 4.7 + Log(1)

= 4.7 + 0

= 4.70

Sample 4.

[Acetic acid] =0.1*5/50 = 0.01 M

[sodium acetate] = 0.1*25/50 = 0.05 M

Now, pH = 4.7 + Log(0.05/0.01)

= 4.7 + Log(5)

= 4.7 + 0.699

= 5.40

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