Would it be possible to show how to calculate the molarity of the Ag ion as well

Question

Would it be possible to show how to calculate the molarity of the Ag ion as well as the Ksp for silver acetate based on my experiment results.
We are doing labs on stuff we haven't even been taught so kinda lost. ncrhle solution again uu, bu l ally residue remains you A. Determination of Silver Acetate in Distilled Water Obtain approximately 50 ml of saturated silver acetate (in distilled water. Rinse the buret with small portions of the silver acetate solution; then fill the buret and record the initial reading to the nearest 0.02 ml (reading the bottom of the meniscus). Pipet Or of the standard 0.050M KCI solution into a cleanEdenmeyer nask. Then add 15 ml of distilled water. Add approximately 1 ml of 5% potassium chromate solution. The color change ofthis indicator is rathersubt dd i indicator is rather subig, from brighn ellaw to a diry" yellow/browo from bright lemon yellow to a "dirty yellow/brown color. Titrate the KCI solution with the silver acetate until the indicator changes color. Record the final volume of silver acetate solution in the buret. Repeat the determination again for consistent results From the concentration of the KCland from the pipet and buret volumes, calculate the concentration, in moles per liter, of silver ion in the saturated solution or silver acetate in istilled water. calculate the solubility product of silver acetate.

Explanation / Answer

Let us first write down the equation that is going on in the Erlenmeyer flask.

AgAc + KCl ------> AgCl (ppt) + KAc

We have employed 10 mL of 0.0500 M KCl solution. Let us estimate the moles of KCl that this 10 mL solution contains.

Amount of KCl used in mole = (10 mL). (1 L/ 1000 mL). (0.0500 mole/L) =

5.0 * 10-4 mole.

From the reaction, we see that KCl and AgAc react on a molar ratio of 1:1, i.e, 1 mole of KCl reacts with 1 mole of AgAc.

Hence, using this argument, we can say that 5.0 * 10-4 mole of KCl (that’s the amount of KCl we added with the pipette) will react with

5.0 * 10-4 mole AgAc.

So, we must have added 5.0 * 10-4 mole of AgAc from the burette. We will need to calculate the molarity of this AgAc solution first.

Volume of AgAc added from the burette

Initial volume (mL)

Final volume (mL)

Volume added (mL)

7.02

14.75

7.73

14.75

22.85

8.10

So, now that we have found out the volume of AgAc added from the burette, we can easily calculate the molarity of the AgAc solution (since we already know the number of moles of AgAc that has reacted). Let us use the first volume (7.73 mL).

Now, let us consider the dissociation of AgAc as follows:

AgAc ------> Ag+ (aq) + Ac- (aq)

We started with 5.0 * 10-4 mole of AgAc, but the concentration of Ag+ in the flask will be influenced by the total volume of the solution. From the first titration, the total volume is (10 + 15 + 7.73) = 32.73 mL. The molar concentration of silver ion is given by {(5.0*10-4 mole)/(32.73 mL)}.{(1000 mL/1 L)} = 0.0153 M.

The concentration of Ag+ in the second titration is 0.0151 M.

The mean molar concentration of Ag+ is (0.0153 + 0.0151)/2 = 0.0152 M (ans).

The same will be the concentration of the acetate ion.

The solubility product is given by

Ksp = [Ag+].[Ac-]

or, Ksp = (0.0152).(0.0152) = 2.31 x 10-4 (ans)

Initial volume (mL)

Final volume (mL)

Volume added (mL)

7.02

14.75

7.73

14.75

22.85

8.10

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