1st table: Standardization Data Table 2nd table: Equivalent Mass Data Table 3rd

Question

1st table: Standardization Data Table

2nd table: Equivalent Mass Data Table

3rd table: pKa data table

1.7

Using the data above answer the following questions:

1. From the standardization data, calculate the molarity of the sodium hydroxide for each trial. Average the values and enter the average in the Standardization Data Table.

2. From the equivalent mass data, calculate the equivalent mass of the unknown acid for each trail. Average the values and enter the average in the Equivalent Mass Data Table.

3. Why is equivalent mass determined and not molar mass?

4. Why must the KHP and the acid samples be dried? If they are not dried, how would the results change (high or low)?

5. Why must NaOH be standardized? Why can't an exact solution of NaOH be prepared?

6. From the graph of pH versus volume of NaOH, determine the pKa of the unknown acid. Convert this value to Ka.

7. Why is the equivalence point in the titration of the unknown acid with sodium hydroxide not at pH 7?

Trail 1 mass KHP, g .44 final volume, mL 36.10 initial volume, mL 11.2 volume of NaOH added, mL 24.9

Explanation / Answer

1. moles KHP = 0.44 g/204.22 g/mol = 0.00215 moles

moles NaOH reacted = 0.00215 moles

molarity of NaOH solution = 0.00215 moles/0.0249 L = 0.08653 M

2. moles NaOH added = 0.08653 M x 55.30 ml = 4.78511 mmoles

moles acid reacted = 4.78511 mmol

equivalent mass of acid = 0.36 g/4.78511 x 10^-3 moles = 75.23 g/mol

3. Equivalent mass is corresponding to number of reactive H's in the acid and thus 1 mole of NaOH reacting with 1 mole of acid here. We can easily find equivalent mass of acid.

4. KHP if not dried would have additional water which would give lower actual moles of KHP in solution than calculate. Thus, volume of NaOH used would be lower, molarity of NaOH thus calculate would be much higher than the true value.

5. NaOH must be standarized as acid concentration is unlnown and to determine the absolute acid concentration in solution we must know how much of base has reacted. moles of base = moles of acid present.

6. From the Part III, pKa = pH at 1.2 equivalence point

pKa = 3.56

7. The weak acid after neutralization (at equivalence point) forms conjugate base A-, which hydrolyze and produces HA and free OH- in solution. Thus the pH of solution at equivalence point is greater than 7 (basic).

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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