Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

4. Part A. The mass of KHC8H4O4 measured to the nearest milligram; however, the

ID: 501853 • Letter: 4

Question

4. Part A. The mass of KHC8H4O4 measured to the nearest milligram; however, the volume of water in which it is dissolved is never of concern-water is even added to the wall of the Erlenmeyer flask during the titration. Explain why water added to the KHC8H4O4 has no effect on the data, whereas water added to the NaOH solution may drastically affect the data.

5. Part B.2. The wall of the Erlenmeyer flask is occasionally rinsed with water from the wash bottle during the analysis of the aid solution. Will this technique result in the molar concentration of the acid solution being reported as too high, too low, or unaffected?

6. Parts A.6. and B.2. For the standardization of NaOH solution in Part A.6. the endpoint was consistently reproduced to a faint pink color. However, the endpoint for the titration of the acid solution in Part B.2 was consistently reproduced to a dark pink color. Will the reported molar concentration of the acid solution be too high, too low, or unaffected by the differences in the colors of the endpoints. Explain.

Explanation / Answer

4) In the experiment we have taken a known mass of KHC8H4O4, Hence we know the number of moles taken

Moles = Mass /Molecular weight

Additon of water will not affect the number of moles.

While in case of NaOH we are using NaOH as solution with a given concentration (moles / Litre), so if we will add water it will affect the concentration. And we will need more of NaOH to titrate the other solution

5) In the flask the moles of titrant are reacting with moles of titrand. So on adding water to flask the number of moles of reactant will remain the same hence the water will not affect the reaction.

6) the dark pink colour shows that we have added excess of NaOH to the flask. So

M1V1 = M2V2

M1 = molarity of base

V1 = Volume of base

M2 = molarity of acid

V2 = Volume of acid

So

M2 = M1V1 / V2

so addition of extra volume of base, the concentration of acid will appear to be too high.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote