CHEM 210 Midterm 2 Exam Questions and Answers All Correct

CHEM 210 Midterm 2 Exam Questions and Answers All Correct Identify the equivalence point on the titration curve. Define the end point of a titration. - Answer-The equivalence point of a titration is reached when the moles of titrant added are stoichiometrically equal to the moles of analyte present in solution. This is observed on the titration curve as the point in the center of the steepest part of the curve, which correlates to the greatest concentration (pH) change per volume of added titrant for the titration. The equivalence point of a titration is the theoretical result of a titration, but what is actually measured is the end point. The end point is when a sudden change occurs in a physical property of the analyte solution that implies equivalence or thereabouts. A common example is the color change of an indicator. In the Fajans titration of Hg2+2Hg22+, NaClNaCl is added to produce the precipitate Hg2Cl2Hg2Cl2. The end point of the titration is detected with bromophenol blue. What charge do you expect the precipitate to have after the equivalence point? - Answer-Negative In the Fajans titration of Hg2+2 with NaCl, the solution contains excess Hg2+2 before the equivalence point. The excess Hg2+2 adsorbs onto the surface of the precipitate, imparting a positive charge onto the precipitate. After the equivalence point, there is excess Cl−Cl− in solution, which adsorbs to the surface of the precipitate, imparting a negative charge onto the precipitate. Describe how the Volhard method can be used to determine the CN−CN− concentration of a solution by placing the steps in the order they occur. - Answer-In the Volhard method, a known excess of standard AgNO3AgNO3 is added to the CN−CN− in 0.5 M HNO3HNO3 solution while stirring vigorously, resulting in the precipitation of AgCNAgCN. Ag+(aq)+CN−(aq)⟶AgCN(s)Ag+⁢(aq)+CN−⁢(aq)⟶AgCN(s) Vigorous stirring is required to prevent excess Ag+Ag+ from becoming trapped in the precipitate as it forms. The next step is to filter off the AgCNAgCN precipitate and wash it with dilute HNO3HNO3. This step is required because AgCNAgCN is more soluble than AgSCNAgSCN, which is formed later in the titration. If AgCNAgCN is not removed from the solution, the end point will slowly fade as AgCNAgCN redissolves and is replaced by AgSCNAgSCN. After the precipitate is removed, Fe(NO3)3Fe(NO3)3 solution is added to the filtrate to give a concentration of 0.2 M Fe3+Fe3+. The excess Ag+Ag+ in the filtrate is then titrated with a standard KSCNKSCN solution. Ag+(aq)+SCN−(aq)⟶AgSCN(s)Ag+⁢(aq)+SCN−⁢(aq)⟶AgSCN(s) Once all of the Ag+Ag+ has been consumed, SCN−SCN− reacts with Fe3+Fe3+ to form a red complex, which signifies the end point of the back titration. Fe3+(aq)+SCN−(aq)⟶FeSCN2+(aq)Fe3+⁢(aq)+SCN−⁢(aq)⟶FeSCN2+⁢(aq) The amount of SCN−SCN− required for the back titration tells you the excess amount of Ag+Ag+ from the reaction with CN−CN−. Because the total amount of Ag+Ag+ added to the solution is known, the amount that reacted with CN−CN− can be determined.