Question

If 38.5 mL of lead (II) nitrate solution reacts completely with excess sodium iodide to yield 0.0628 g or precipitate, what is the molarity of lead (II_ ion in the original solution.

I was wondering if any could tell me how to solve this I have been looking over my chapter and can't figure out what I need to do.

Answer 1

Might be best to first write up the Chemical formulas of each compound first.

Lead II Nitrate = Pb(NO3)2
Sodium Iodide = NaI

Then predict the products.

Once the products are correct, balance the chemical equation.

Then you have to realize which of the products is the precipitate (insoluble), convert 0.0628 grams of this product into moles, then convert the moles of this product to moles of lead II nitrate (in the reactants side), convert the moles of lead II nitrate to moles of the lead II ion, and finally....

Divide moles of lead II ion by liters of solution to give concentration in molarity.

Answer 2

How do you convert lead (II) nitrate to lead (II) ions?

Answer 3

it's sort of trivial in this case: for every Pb(NO3)2 molecule in solution, there are two NO3- and one Pb2+ ion; so it's a 1:1 conversion

Answer 4

So if there is 1.88 x 10^-4 Pb(NO3)2 then there is 1.88 x 10^-4 Pb ions

Answer 5

that's true, but where did you get these numbers? you have to first predict the products, which are PbI2 (s) and NaNO3 (aq) and balance the equation. then you would know that since you have 0.0628 g of precipiate (read: solid), the precipitate must be PbI2. then convert 0.0628 g of PbI2 to moles of PbI2. did you get this far?

Answer 6

I have set up the reaction and balanced it I have also found the moles of PbI since there is a 1:1 ratio for Pb(NO3)2 to PbI2 then 1.36x10^-4 (sorry re-did the math was little off) of Pb(NO3)2 equals 1.36x10^-4 of PBI2. Am I correct so far?

Answer 7

Yes, pretty much. What's next?

Answer 8

Since there is a 1:1 ratio of Pb(NO3)2 to Pb ions then there are 1.36x10^-4 of Pb ions divide by .0385L=0.00538 how am I doing.

Answer 9

(1.36e-4)/(.0385) should equal 3.53e-3 M or 3.53 mM

Answer 10

oops it is 3.54x10^-3 though correct? Does this work for every reaction?

Answer 11

That probably is correct.

The principles will always work, but you have to know how to apply them properly toward your goal.

For instance,

Molarity will always be # moles / # of Liters of solution.

Converting between compounds in a reaction will always work if you convert everything to moles first and use the coefficients properly (in this case it was negligible, since everything was 1:1).

And mixing two soluble ionic compounds in an aqueous solution will always result in an exchange of ions (each anion pairs with the other cation).

Answer 12

Thank you very much

Answer 13

np.