I'm stuck on this one aswell...
50.0 ml of a 0.0500 M solution of lead (II) nitrate is mixed with 40.0 ml of a 0.200 M solution of sodium iodate at 25°C. Calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium. At this temperature the Ksp for lead (II) iodate is 2.6 x 10-13
Write the balanced equation
Pb(NO3)2 + 2NaIO3 ==> Pb(IO3)2 + 2NaNO3
This is a Ksp problem with a common ion. You will need to determine the mols of each reactant from M x L = mols, see how much of the product is formed and which reactant is in excess. After you understand how to do a and b of the previous post you will know how to do this one.
To calculate the Pb2+ and IO3- concentrations when the mixture comes to equilibrium, we need to use the balanced equation and the concept of molarity.
First, let's calculate the number of moles (mols) of each reactant. We can use the formula M x L = moles, where M is the molarity in moles per liter and L is the volume in liters.
For the lead (II) nitrate (Pb(NO3)2):
Moles of Pb(NO3)2 = 0.0500 M x 0.0500 L = 0.00250 mol
For the sodium iodate (NaIO3):
Moles of NaIO3 = 0.200 M x 0.0400 L = 0.00800 mol
Now, let's determine the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the amount of product formed.
From the balanced equation, we can see that the mole ratio between Pb(NO3)2 and NaIO3 is 1:2. This means that for every 1 mol of Pb(NO3)2, 2 moles of NaIO3 are required. Since we have 0.00250 moles of Pb(NO3)2 and 0.00800 moles of NaIO3, we can see that Pb(NO3)2 is the limiting reactant.
Next, let's determine how much of the product, Pb(IO3)2, will form. Since the mole ratio between Pb(NO3)2 and Pb(IO3)2 is 1:1, we know that the moles of Pb(IO3)2 formed will be equal to the moles of Pb(NO3)2.
Therefore, the moles of Pb(IO3)2 formed = 0.00250 mol
Now, we can convert the moles of Pb(IO3)2 into molarity by dividing by the total volume of the mixture, which is the sum of the initial volumes for Pb(NO3)2 and NaIO3:
Total volume = 50.0 mL + 40.0 mL = 90.0 mL = 0.0900 L
Concentration of Pb2+ = moles of Pb(IO3)2 / total volume
Concentration of Pb2+ = 0.00250 mol / 0.0900 L = 0.0278 M
Since the mole ratio between Pb(IO3)2 and IO3- is 1:2, we know that the moles of IO3- formed will be twice the moles of Pb(IO3)2 formed:
Moles of IO3- formed = 2 x 0.00250 mol = 0.00500 mol
Concentration of IO3- = moles of IO3- / total volume
Concentration of IO3- = 0.00500 mol / 0.0900 L = 0.0556 M
Therefore, the concentrations of Pb2+ and IO3- when the mixture comes to equilibrium are 0.0278 M and 0.0556 M, respectively.