Question

Use the following graph to answer questions 3-6. 160.0 40.0 80.0 120.0 ml. of 0.100 M NaOH added 3. The pKa, of this weak acid is about A. B. 5.0 10.0 C. D. 3.8 11.0 4. The pKof this weak acid is about A. B. 5.0 9.0 C. 3.8 D. 11.0 5. The 2nd equivalence point of this acid occurs at approximately A. 40 mL B. 60 mL C. 80 mL D. 100 mL 6. A buffer will occur at which points on the graph?

Answer 1

3)The given graph has 2 jumps (a sudden increase in pH value on adding only a little quantity of the base) and so it has 2 equivalence points.

Let us consider the first equivalence point,i.e.,the first jump in the curve.It occurs from pH 6 to pH 10. We will consider the mid point of this jump,which is referred to as the equivalence point.It corresponds to 40 ml of NaOH. Now the half of 40 ml is 20 ml. A vertical straight line is drawn from the point corresponding to 20 ml upto the titration curve.Then a horizontal straight line is drawn from the titration curve to the pH axis.The point where this line touches the pH axis,gives the value of pH of the solution. The pH corresponding to 20 ml of NaOH, i.e. the half of the equivalence point is the value of  pKa1 .It corresponds to the pH 5.0. So, pKa1 = 5.0 (option a) .

4) For the second jump,the equivalence point corresponds to nearly 80 ml of NaOH.The 3/4 th of 80 ml is 60 ml.The pH corresponding to 60 ml of NaOH , i.e. the 3/4 th of the equivalence point is the value of pKa2.Thus, pKa2 = 9 (option b).

5)The second equivalence point is the mid point of the second jump in the curve.It corresponds to approximately 80 ml (option c).