Water pipes, especially older ones, often contain lead. One example of lead prob
ID: 517894 • Letter: W
Question
Water pipes, especially older ones, often contain lead. One example of lead problems was in Flint, Michigan, where a change in the drinking water source led to a high concentration of lead ion in their drinking water. a. Solid lead is converted to lead (II) ion by oxidation with dissolved O_2 in the water Write. Write a complete balanced chemical reaction for the oxidation of lead to lead (II) in an acidic environment. b. The ratio of chloride to sulfate ions contributes to the quantity of lead ion in the water. Using the published K_sp values in Appendix C, would you want more chloride or sulfate ions present in the water to minimize the amount of dissolved lead (II) ion? Explain. c. Based on the K_sp value you calculated for PbI_2, would it be better to add iodide, chloride, or sulfate to a water supply to decrease the concentration of lead ions? Why?Explanation / Answer
The redox reaction of plumber or lead is the next equation.
2Pb0 + O2 = 2PbO
This will be the balanceated equation. If we balance this equation by multipliyng the reaction by 2 the atoms will be the same in the reactants and the products.
Another form to solve this will be balance the reaction with a redox method.
The reaction will be tje next one.
Pb0 = Pb2+ + 2e- To balance the charges we need to put 2 electrons in the products side. With this, both mass and charge is balanced.
Then we need to balance the oxygen.
2H++O2 = O2- + H2O In this case we need to balance the mass.We use molecules of water to balance the reaction. In this case we use one molecule to have two molecules of oxygen in both sides.
The we balance hydrogen. We use protons because the problem needs that. An for that we use two electrons. So in this case we need also balance the charge. If we meka the sum of charges we will see that the entire charges in both sides is zero.
So with that we only ned to write the sum.
Pb0 = Pb2+ + 2e-
2H++O2 = O2- + H2O .........To have the same electrons we multiply 2 electrons in oxygen reaction.
(2H++O2 = O2- + H2O)2 and the result will be:
4H++2O2 = 2O2- + 2H2O So we make the sum.
Pb0+ 4H++ 2O2 +2= 2O2- + 2H2O + Pb2+ + 2e- ........If we make the charges sum wi wll note that the value is zero. And the mass is also balanceated.
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