A chemist takes a 647 gram sample of this hydrate. She removes all the water possible from the sample and collects it. To this water, she adds 7.6 grams of lead(II) nitrate. Afterwards she wants to precipitate all the lead. She wants to use a solution of aluminum sulfate to precipitate the lead. The aluminum sulfate solution is 0.07 M and she has 1.0 L of it.

a. What is the Molarity of the lead(II) nitrate solution ?
b. How many ml of aluminum sulfate solution will the chemist need?
c. What will be the resulting Molarity of the aluminum cation left in the resulting solution mixture once she combines her volumes ?

John, I don't believe you have all of the information listed. You can do b but not a or c with the information listed.

Oops, the hydrate is Copper Sulfate Pentahydrate

To find the answers to these questions, we will use the concept of stoichiometry and molarity.

a. The molarity (M) of a solution is defined as the number of moles of solute divided by the volume of the solution in liters. To find the molarity of the lead(II) nitrate solution, we need to first calculate the number of moles of lead(II) nitrate.

First, let's convert the mass of lead(II) nitrate to moles. The molar mass of lead(II) nitrate (Pb(NO3)2) is 207.2 g/mol (lead) + (2 * 14.01 g/mol (nitrogen)) + (6 * 16.00 g/mol (oxygen)) = 331.2 g/mol.

The number of moles of lead(II) nitrate = mass / molar mass = 7.6 g / 331.2 g/mol = 0.0229 mol.

Now, to find the molarity, we divide the number of moles by the volume of the solution in liters (which is not given). Since we don't have the volume, we cannot determine the molarity of the lead(II) nitrate solution without that information.

b. To determine how many milliliters of aluminum sulfate solution the chemist will need, we need to use stoichiometry.

First, let's write the balanced chemical equation for the reaction between lead(II) nitrate (Pb(NO3)2) and aluminum sulfate (Al2(SO4)3):

3 Pb(NO3)2 + 2 Al2(SO4)3 → 3 PbSO4 + 2 Al(NO3)3.

From the balanced equation, we can see that 3 moles of lead(II) nitrate react with 2 moles of aluminum sulfate to form 3 moles of lead sulfate and 2 moles of aluminum nitrate.

Now, let's determine the number of moles of lead(II) nitrate. We already calculated it earlier; it is 0.0229 mol.

Using stoichiometry, we can see that 3 moles of lead(II) nitrate react with 2 moles of aluminum sulfate. Therefore, the ratio of moles of lead(II) nitrate to moles of aluminum sulfate is 3:2.

To calculate the number of moles of aluminum sulfate, we can use the following proportion:

(0.0229 mol Pb(NO3)2) / (3 mol Pb(NO3)2) = (x mol Al2(SO4)3) / (2 mol Al2(SO4)3),

where x is the number of moles of aluminum sulfate.

Solving for x, we find x = (0.0229 mol Pb(NO3)2 * 2 mol Al2(SO4)3) / (3 mol Pb(NO3)2) = 0.01526 mol Al2(SO4)3.

Now, let's determine the volume of aluminum sulfate solution needed. The molarity of aluminum sulfate solution is 0.07 M, which means there are 0.07 moles of aluminum sulfate per liter of solution.

Using the given volume of 1.0 L, we can calculate the number of moles of aluminum sulfate:

0.07 mol/L * 1.0 L = 0.07 mol.

To find the volume of aluminum sulfate solution needed, we can use the molarity equation again:

molarity = moles of solute / volume of solution in liters.

Rearranging the equation, we have:

volume of solution = moles of solute / molarity,

volume of solution = 0.01526 mol Al2(SO4)3 / 0.07 mol/L = 0.218 L = 218 mL.

Therefore, the chemist will need 218 mL of aluminum sulfate solution.

c. To determine the resulting molarity of the aluminum cation left in the resulting solution mixture once the volumes are combined, we need to calculate the total moles of aluminum cation and the total volume of the mixture.

Since we are combining 1.0 L of the aluminum sulfate solution, the final volume of the mixture will be 1.0 L.

The total amount of aluminum cation can be calculated using the number of moles of aluminum sulfate we determined earlier:

0.01526 mol Al2(SO4)3 * (2 moles Al3+/1 mole Al2(SO4)3) = 0.03052 mol Al3+.

The resulting molarity is given by:

molarity = moles of solute / volume of solution in liters.

So, the resulting molarity of the aluminum cation will be:

molarity = 0.03052 mol Al3+ / 1.0 L = 0.03052 M.

Therefore, the resulting molarity of the aluminum cation left in the resulting solution mixture will be 0.03052 M.