Question

I need help with part 6! All the other questions are correct so far! Thank you!

In Part 1 of Lab 2 you will make and standardize a solution of NaOH(aq). Suppose in the lab you measure the solid NaOH and dissolve it into 100.0 mL of water. You then measure 0.1995 g of KHP (KC₈H₅O₄, 204.22 g/mol) and place it in a clean, dry 100-mL beaker, and then dissolve the KHP in about 25 mL of water and add a couple of drops of phenolphthalein indicator. You titrate this with your NaOH(aq) solution and find that the titration requires 9.75 mL of NaOH(aq)

Part 1:

What is the concentration of your NaOH(aq) solution?

1.00×10^-1 M NaOH

Part 2:

Suppose you then make a Kool-Aid solution by dissolving the powder in an entire packet of lemon-lime Kool-Aid in 250.00 mL solution.  You find the powder in a packet of lemon-lime Kool-Aid has a mass of 3.639 g.

In Part 2 of the lab you titrate 5.00 mL of the lemon-lime Kool-Aid (with 5 drops of thymol blue indicator) with your NaOH(aq) solution. The titration requires 10.32 mL of NaOH(aq).

Determine the number of moles of NaOH(aq) required to titrate 5.00 mL of your Kool-Aid solution.

1.03×10^-3 mol NaOH

Part 3:

Determine the number of moles of NaOH(aq) that would be required to titrate 250.00 mL of your Kool-Aid solution (the entire packet).

5.25×10^-2 mol total acid

Part 4:

In Part 4 of the lab you titrate 25.00 mL of the lemon-lime Kool-Aid (with KI, HCl, and starch) with 0.001000 M KIO₃(aq) solution. The titration requires 10.64 mL of KIO₃(aq).

Determine the number of moles of ascorbic acid in 25.00 mL of your Kool-Aid solution.

3.192×10^-5 mol ascorbic acid

Part 5:

Determine the number of moles of ascorbic acid in 250.00 mL of your Kool-Aid solution (the entire packet).

3.192×10^-4 mol ascorbic acid

Part 6:

If you titrated the entire packet of Kool-Aid, how many moles of NaOH(aq) would be neutralized in the titration due to the ascorbic acid?

Answer 1

Dear Student,

I am assuming that answer of part 1 to part 5 is correct. Kool- Aid is an acidic mixture of HCl and Ascorbic acid. In part 3 you have calculated total number of moles of acid present in the mixture, that is 5.25 x 10-2. This number is sum of number of moles of HCl and ascorbic acid. In part 5 you have calculated total number of moles of ascorbic acid present in 250 ml solution. This many number of moles, 3.192 x 10-4, of NaOH will be required to neutralize ascorbic acid alone as it is a monobasic acid. So 3.192 x 10-4 moles out of 5.25 x 10-2 moles of NaOH ( total number of moles required to neutralize the mixture) will be neutralized due to ascorbic acid.

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