Results Volume 0.200 M NaOH in step (1)= 50mL Volume 0.200 M NaOH in step (2)= 5

Question

Results

Volume 0.200 M NaOH in step (1)= 50mL

Volume 0.200 M NaOH in step (2)= 50mL

Volume 0.200 M NaOH in step (3)=75 mL

Calculate

1- Calculate the concentration of acetic acid (in moles/L)in the original uncatalyzed reaction mixture based on step(1) results.

2- Calculate the concentration of acetic acid (in moles/L)in the mixture that has reached equilibrium, being sure to apply the correction in volume of NaOH used based on the sulfuric acid added based on step (2) results.

3- Based on your results in 1 and 2 complete the following table

4- Calculate the equilibrium constant Kc for the reaction being sure to include the concentration of water (since this is not a dilute aqueous solution).

part 2

Why is it necessary to experimentally determine the molarity of the NaOH solution used in the titrations in this experiment instead of just using the value calculated by your dilution?

Part 3

1- Write the reaction and the Ksp expression for the following slightly soluble salts dissolving in water.

(a). Lead Iodide (PbI2)

(b). Calcium Iodate Ca(IO3)2

2- A student performed the titrations done in this experiment. The student pipetted 25mL of saturated calcium hydroxide solution into a flask and then titrated it with 0.103M HCl. It required 22.7mL of HCl to reach the orange-red point. Calculate the molar solubility and Ksp of calcium hydroxide.

3- A student performed the titrations done in this experiment. The student pipetted 25mL of solution that contained saturated calcium hydroxide solution and 0.05 M calcium ions into a flask and then titrated it with 0.103 M HCl. It required 2.77mL of HCl to reach the orange red end point. Calculate the molar solubility of calcium hydroxide in this solution.

Ice Table Acetic Acid Isopropyl Acetate Water Initial Concentration Change in Concentration Equilibrium concentration

Explanation / Answer

In the first part, I need the volume of acetic acid, so I can calculate that.

For part 3, I'll answer question 1 and 2. Question 3 you can answer it in a similar way to answer question 2:

1. PbI2(s) ---------> Pb2+ + 2I- Ksp = [Pb2+] [I-]2

Ca(IO3)2(s) ------------> Ca2+(aq) + 2IO3-(aq) Ksp = [Ca2+] [IO3-]2

2. With this data, all you need to do is apply the following equation:

MaVa = MbVb

Ca(OH)2 + 2HCl --------> CaCl2 + H2O

0.103 * 22.7 = 2 * 25 * Mb

Mb = 0.047 M = solubility.

Ca(OH)2 -------> Ca2+ + 2OH-

Ksp = [Ca] [OH]2

Ksp = (s) (2s)2

Ksp = 4s3

Ksp = 4 * (0.047)3

Ksp = 4.15x10-4

Question 3 is the same but you have to add the 0.05 M to the [Ca] in the expression of Ksp.

Provide the missing data of part 1 and post it in a new question.

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