Question

A student is trying to calculate the unknown concentration of a lead nitrate solution Pb(NO3)2. Potassium iodide, KI reacts with the lead nitrate to form a precipitate of Lead Iodide PbI2. If 10.0 ml of lead nitrate is titrated with 38.7 ml of KI to form 12.78 g of lead iodide.

a. Write a brief description of a procedure that can be used to obtain the above data.
b. Calculate the following:
i. The moles of lead iodide formed.
ii. The moles of lead nitrate in the 10.0 ml sample.
iii. the concentration in mol/L of lead nitrate solution.
c. Write the valance electron configuration of both Lead and Iodine.
d. Use the electron configuration to describe the electrons that are transferred to iodine to form the compound PbI2.
e. Lead will form two different compounds with PbI2 and PbI4, using lead's electron configuration explain how PbI4 would form.

Answer 1

To obtain the given data, a procedure could involve titrating lead nitrate with potassium iodide and collecting the precipitated lead iodide. The moles of lead iodide formed can be calculated using the given mass and molar mass. The concentration of lead nitrate can be calculated using the moles and volume of the sample.

Step-by-step explanation:

To obtain the given data, a procedure could involve titrating 10.0 mL of lead nitrate solution with 38.7 mL of potassium iodide solution and then collecting the precipitated lead iodide. The procedure should be performed in a controlled environment and under proper safety measures.

In order to calculate the moles of lead iodide formed (part b.i), you can use the given mass of lead iodide (12.78 g) and its molar mass (461.01 g/mol) to calculate the moles of the compound.

The moles of lead nitrate in the 10.0 mL sample (part b.ii) can be calculated using the volume of the sample and the concentration of the lead nitrate solution.

The concentration in mol/L of the lead nitrate solution (part b.iii) can be calculated using the moles of lead nitrate and the volume of the sample.

The valence electron configuration of lead is [Xe] 4f^14 5d^10 6s^2 6p^2, and the valence electron configuration of iodine is [Kr] 4d^10 5s^2 5p^5 (part c).

In the formation of PbI2, lead transfers its 2 valence electrons to iodine to satisfy the octet rule (part d).

To form PbI4, lead would need to have an expanded octet. However, based on its electron configuration, it is not favorable for lead to form PbI4.