Question

5. The Ka and Molar Mass of a Monoprotic Weak Acid

a. Suppose that–unknown to you–the primary standard KHP (potassium hydrogen phthalate, KHC8H4O4) had a potassium iodide impurity of approximately one percent by mass. How would this have influenced the calculated molarity of your sodium hydroxide solution? Would your calculated value be too low, too high, or unchanged? Explain your answer.

b. Sketch a typical titration curve for a monoprotic weak acid titrated with a strong base. Label the axes to show the equivalence volume and the neutral point (where pH = 7).

c. For the titration of 0.699 g of an unknown monoprotic weak acid, a 39.87-mL volume of a 0.1002M NaOH solution was required to reach the equivalence point. Calculate the molar mass of the unknown weak acid.

Answer 1

Monoprotic acid (a) Molarity of NaOH is calculating by titration the primary standard KHP Even KHP is impure we considered the whole mass weight is corresponding to KHP. So, the number of moles calculate from its mass is greater than actually number of moles of KHP The moles of NaOH can be calculated by equating the number of moles of KHP. So, the number of mole s of NaOH getting is greater than the number of moles ofNaOH actually present. Since the number of moles per volume is molarity, we get the molarity value of NaOH to be too high (b) Weak acid solution pH is in between 4 to 6.5 (nearly) and the strong base solution pH is above 10. Before equivalence point the pH of titration is below 6 and at equivalence point, pH is suddenly rise to above 10. At half-the equivalencepoint the pKa of weak acid is equal to the pH of the titration. 14 12 10 pH Equivalenced point pKa Volume of Strong base Ve is the equivalence volume of strong base (c) The number of moles of Acid - The number of moles of NaPH (3987 mL x 100 LL-(01002 M) -39.87 mLX 1000 mL 0.003995 mol 0.699g 175 g/mol 0.003995 mol mass of acid Molar mass of acid - moles of acid