Question

7. (20 pts) A scientist has 100 mL of a solution that contains four metal ions, Ca2, Cu2, Pb2, and K*. He takes 50 mL and adds 2.5 grams of Na2SO4 which precipitates the Ca2 and the Pb2. This precipitate is filtered form the solution and redissolved in 20 mL of 6 M HNO3. He then adds 20 mL of 1.005 M EDTA and back titrates with 23.54 mL of 0.1954 M Zn2 He then adds a masking agent that complexes with the Pb2, and it takes 46.06 mL of the 0.1954 M Zn2 to reach the endpoint. He knows that he can analyze for the Cu2 spectroscopically. So he makes a calibration curve and records the following data: M Cu2+ Abs 0.005 0.068 0.015 0.203 0.025 0.035 0.342 0.488 He records the absorbance of the original solution to be 0.401. He runs 20 mL of the original solution through a column that only allows groupl ions to pass through. He titrates the sample that he collects with 20.4 mL of 0.225 M EDTA. Determine the mole percent of each ion present in the original sample.

Answer 1

The Absorbance of the Original solution = 0.400. from Calibration Curre, he records following M a 2t Absorbance 0.068 0.005 0.015 0.203 0.025 0: 342 1 0.035 0.488 Therefore, the absorbance value for cutt, molarity of oogs is 48.0.342 Which is very close to the original value 0.401. However, Original solution lies between 0.025 and 0.035.

M cu ² of 0.025 Contains 1.988 g of cut m cu 27 of 0.035 Contains 2.7848 of cult Explaination. Molarity = mass molar mess 0.025= mass (9) 10.035= mass (M) 79.51 (Soles) 79.545 g/mol Mass (g) = 79.54580.025 Mass= 79.54580035 Cu = 2.784 9 wom rcs 1.988 9 10 Original solection absorbance can be obtained by trial and error method. J We know that - for M, Cu 2+ for 0.005 is 0.068 absorbance and for M, cu 2 of 0.02s is 0.342 alsabene we required, 0.401 I 'M , curt of 0.09 may be having absorbere & 0.401.

Absorbance, A= {cl. Beer - lamber law. Absorbance, totally depends on Concentration because the absortiving & length is constant throughout