2.) If you pass 10.0mL of a 50.0mL solution of [Co(en)2Cl2]Cl that has been reduced with Zn through a cation exchange column, you will obtain a solution that has three H+ ions for every Co3+ ion that was originally present in the sample. These H+ ions are titrated with 0.10-M NaOH solution. It is found that 20.0mL of NaOH are required. What volume of .1M agNO3 solution is required to precipitate all the chloride as AgCl?

Question

2.) If you pass 10.0mL of a 50.0mL solution of [Co(en)2Cl2]Cl that has been reduced with Zn through a cation exchange column, you will obtain a solution that has three H+ ions for every Co3+ ion that was originally present in the sample. These H+ ions are titrated with 0.10-M NaOH solution. It is found that 20.0mL of NaOH are required. What volume of .1M agNO3 solution is required to precipitate all the chloride as AgCl?

Answer 1

To find the volume of 0.1 M AgNO3 solution required to precipitate all the chloride ions as AgCl, we need to understand the stoichiometry of the reaction between silver nitrate (AgNO3) and chloride ions (Cl-).

The balanced equation for the reaction between AgNO3 and Cl- is:
AgNO3 + Cl- → AgCl + NO3-

From the information given in the question, we know that the [Co(en)2Cl2]Cl solution is reduced by Zn, forming three H+ ions for every Co3+ ion. However, this information is not directly relevant to finding the volume of AgNO3 solution required for precipitating chloride ions.

We can assume that the [Co(en)2Cl2]Cl solution is a source of chloride ions (Cl-), which will react with AgNO3 to form AgCl precipitate.

To calculate the volume of AgNO3 solution required, we can use the concept of stoichiometry.

First, we need to determine the number of moles of chloride ions present in the 10.0 mL of [Co(en)2Cl2]Cl solution passed through the cation exchange column. We can use the formula:

moles of chloride ions = concentration of chloride ions × volume of solution

The concentration of chloride ions can be calculated from the molarity of the [Co(en)2Cl2]Cl solution. Assuming that the 50.0 mL solution has a molarity of x M, we have:

moles of chloride ions = x M × 10.0 mL

Next, we need to determine the stoichiometry of the reaction between AgNO3 and Cl-. From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl-. Therefore, the number of moles of AgNO3 required will be equal to the number of moles of chloride ions present.

Now we can calculate the volume of 0.1 M AgNO3 solution required to react with the moles of chloride ions, using the formula:

volume of AgNO3 solution = moles of chloride ions / molarity of AgNO3

Substituting the values into the equation, you can calculate the volume of 0.1 M AgNO3 solution required to precipitate all the chloride as AgCl.