A. Using a 500 mL volumeteric flask, prepare 500 mL of 0.1 M NaOH solution from

Question

A. Using a 500 mL volumeteric flask, prepare 500 mL of 0.1 M NaOH solution from a 6 M NaOH stock solution.

i. Add ` 300 mL of DI H2O to a 50 0mL volumetic flask.

ii. caluclate the amount of 6M NaOH needed to make a 0.1 M NaOH solution when diluted to a total volume of 500 mL.

iii.Measure out this amount of 6 M NaOH with a graduated pipet and place it in a 500 mL volumetic flask.

iv. Add more DI H2O to the volumetic flask slowly until the meniscus is at the line of the volumetiric flask

B. weight out approx 0.3 g of KHP.

C. Add each sample of the KHP to three seperate 100 mL flasks.

D. Add 25 mL of DI H2O to each flask containing KHP(3) and swirl to dissolve.

E. Add a few drops of phenolphthalein to the KHP solution flasks.

I. Standardized of approx 0.1 NaOH Solution

1. Mass of KHP Samples (include units)

Sample 1: 0.293

Sample 2: 0.280

Sample 3: 0.279

2. Molar Mass of KHP: 204.22

3. Calulcate the number of moles of KHP in each sample (include units)

Sample 1: ? Sample 2: ? Sample 3: ?

4. Record the colume of NaOH required to reach the end point for each sample of KHP.

Sample 1: 17.5 Sample 2:16.8 Sample 3: 24.5

5. At the end point of each titration, how many moles of NaOH have you added to each sample of KHP?

Sample 1: ? Sample 2: ? Sample 3:?

6 . Calculate the molarity of the NaOH solution for each sample.

Sample 1:? Sample 2:? Sample 3: ?

I don't know how to do number 3, number 5 and number 6. Could someone please help me? I would greatly appreciate all of your help. Thank you in advance!

Explanation / Answer

3. Calculate the number of moles of KHP in each sample (include units)

Number of moles = mass of KHP / molecular weight of KHP

So for

Sample 1:

Number of moles = 0.293 g / 204.22 = 0.001435 Moles of KHP

Sample 2:

Number of moles = 0.280 g / 204.22 = 0.001371 Moles of KHP

Sample 3:

Number of moles = 0.279 / 204.22 = 0.001366 Moles of KHP

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5. At the end point of each titration, how many moles of NaOH have you added to each sample of KHP?

In a titration the end point is where the titrating agent and the reagent titrated will be equal in concentration and the solution becomes neutral.

So, at end point in each batch of titration the number of moles of KHP will be equal to that of NaOH.

Hence

Sample 1:

Moles of NaOH = 0.001435 Moles

Sample 2:

Moles of NaOH = 0.001371 Moles

Sample 3:

Moles of NaOH = 0.001366 Moles

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6 . Calculate the molarity of the NaOH solution for each sample.

Molarity of a solution = Number of moles of solute / Solvent Volume in L

Since all samples are made to 500 mL (0.5L) in a volumetric flask molarity of the NaOH solution for each sample can be calculated as follows:

For sample 1 = 0.001435 Moles / 0.5 L = 0.00287 M

For sample 2 = 0.001371 Moles / 0.5 L = 0.002742 M

For sample 3 = 0.001366 Moles / 0.5 L = 0.002732 M

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