Buffers and Potentiometric Analysis lab 1.Assuming that the unknown weak acid so

Question

Buffers and Potentiometric Analysis lab

1.Assuming that the unknown weak acid solution is vinegar (aqueous acetic acid, CH3COOH) with a density of 1.00 g/mL, calculate the mass percent of acetic acid in your unknown solution.

2.If the endpoints you reached during the standardization of your NaOH solution were very precise (relative average deviation of 0.03%) but were all overshot, is the reported molar concentration of the acid solution from this experiment greater than, less than, or equal to the true concentration? Explain your answer.

3. If your absent-minded lab partner spilled deionized water into your NaOH titrant, diluting it, would the Ka value you determined for the weak acid be greater than, less than, or equal to the true Ka for the acid? Explain your answer.

4. In part II of the lab, you are given three buffer solutions. One solution is a very good buffer, while the other two are not very good buffers. The two poor buffers are ineffective buffers for different reasons. Explain why the buffers are not good at resisting pH changes.

Explanation / Answer

Answer to Question 1) Vinegar is an aqueous acetic acid solution. To analyse the amount of acetic acid in the vinegar sample, we need to carry a simple acid-base titration with phenolphthalein indicator. In this titration we need prepare NaOH solution of known concentration, accurately measure the volume of vinegar and record the volume of NaOH required to neutralise the acid.

CH3COOH(aq) + NaOH(aq) -------> CH3COONa(aq) + H2O(l)

Let the volume of NaOH added =A mL or A X 10-3 L

Number of moles of NaOH= concentration of NaOH X volume of NaOH

Number of moles of NaOH= c mol L-1 X A X 10-3 L = B mol

From the above equation we know that 1 mol of acetic acid are consumed per mol of NaOH

Number of moles of acetic acid = B mol

Now we can determine the mass using the following formula

mass of acetic acid = number of moles of acetic acid (n) X Molar mass of acetic acid (M)

mass of acetic acid = B mol X 60.05 g mol-1 = C g

The density of the vinegar sample is 1.00 g mL-1, mass of vinegar is given by:

mass of vinegar = volume of vinegar X density of vinegar

mass of vinegar = D mL X 1.00 g mL-1 = E g

The percent by mass of acetic acid in the vinegar sample is given as:

% acetic acid = (mass of acetic acid / mass of vinegar) X 100 = F %

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.