In the determination of the equivalent weight of an unknown acid, a student took
ID: 476724 • Letter: I
Question
In the determination of the equivalent weight of an unknown acid, a student took a sample weight 1.54g and followed the procedure of this assignment. (Determination of the equivalent weight of an acid) However, he found that he had passed the end point when 15.6ml of his standard base had been added. In order to prevent the loss of the determination, he added 10.00 ml of his standardized sulfuric acid and resumed the titration. The end point was reached when a total of 23.40 ml of base had been added. The concentration of his standardized sodium hydroxide was 0.592N, while that of his standardized sulfuric acid was 0.463N. a. What informatioon, if any could have been obtained about the equivalent weight of the solid acid from the fact that 15.6ml of base had been added at the time the student realized that he has passed the end point? b. Compute the equivalent weight of the solid acid by use of all the data. c. What additional experimental errors were introduced as a result of having passed the first end point?
Explanation / Answer
Let the equivalent weight of the unknown acid be E g/equivalents; number of equivalents of the unknown acid = (1.54 g)/(E g/equiv) = (1.54/E) eq.
Write down the balanced equation for the reaction of NaOH with H2SO4:
H2SO4 + 2 NaOH -------> Na2SO4 + 2 H2O
Since 1 equiv. of H+ from sulfuric acid combines with 1 equiv. of OH- (from NaOH) to form 1 equiv. of H2O, hence,
1 equiv. of H2SO4 = 2 equiv. of NaOH (since H2SO4 is diprotic – contains two protons).
Equivalents of H2SO4 used = (concentration in normality)*(volume in L) = (0.463 N)*(10.00 mL) = (0.463 N)*(10.00 mL)*(1 L/1000 mL) = 0.00463 equiv.
Equivalents of NaOH used for the combined titration = (0.529 N)*(23.40 mL) = (0.529 N)*(23.40 mL)*(1 L/1000 mL) = 0.0123786 equiv.
The sum of the equivalents of the unknown acid and sulfuric acid must equal the equivalents of NaOH used, since the solution is completely titrated. Therefore,
(1.54/E) + 0.00463 = 0.0123786
===> 1.54/E = 0.0123786 – 0.00463 = 0.0077486
===. E = 1.54/0.0077846 = 198.7455 198.75
The equivalent weight of the unknown acid is 198.75 (ans).
When 15.6 mL of standard NaOH were added to the solution of the unknown acid, the equivalents of NaOH in the solution = (0.529 N)*(15.6 mL) = (0.529 N)*(15.6 mL)*(1 L/1000 mL) = 0.0082524; this value is certainly more than the equivalents of the unknown acid (0.0077486) present and hence the solution was over-titrated.