I really don't understand how to set this up. Any help is really appreciated! I really don't quite understand how to set up
a caluculation to figure out whether a precipitate will form and i'm struggling with the second half too!
When 25.75 mL of 0.00826 M lead II nitrate is mixed with 75.10 mL of 0.0183 M sodium chloride solution, will a precipitate of PbCl2 form? Show all calculations to prove your answer. The ksp for lead II chloride is 1.7 x 10^-5.
Just remember what Ksp is!!
Ksp, in this case for PbCl2, = 1.7 x 10^-5. What does that mean? It simply means that (Pb^+2)(Cl^-)2 MAY NOT EVER EXCEED 1.7 x 10^-5. So you calculate the product of those concns, (raise each to the power, of course, in the balanced equation) and if that number is higher than Ksp, a ppt will occur. If that number is less than Ksp, no ppt will occur. If that number is the same as Ksp, then the solution is exactly saturated and the next Pb ion and Cl ion added to that solution will ppt (1 molecule of PbCl2).
mols lead (II) nitrate = M x L = ??
mols NaCl = M x L = ??
(Pb^+2) = mols/L (remember to add mL lead nitrate to mL NaCl to find total mL of solution. We assume the volumes are additive.
(Cl^-) = mols/L.
Then (Pb^+2)(Cl^-)^2 = ?? and compare with Ksp.
Let me know if this isn't clear.
To determine if a precipitate will form when two solutions are mixed, we need to compare the ion concentrations in the combined solution to the solubility product constant (Ksp) of the compound we're interested in.
In this case, we want to find out if a precipitate of PbCl2 will form when lead II nitrate and sodium chloride solutions are mixed.
Step 1: Determine the ions present in the solutions.
Lead II nitrate dissociates into Pb2+ and NO3- ions.
Sodium chloride dissociates into Na+ and Cl- ions.
Step 2: Write the balanced chemical equation for the reaction.
Pb(NO3)2 + 2NaCl --> PbCl2 + 2NaNO3
Step 3: Calculate the moles of lead II nitrate and sodium chloride.
Moles = concentration (M) x volume (L)
Moles of Pb(NO3)2 = 0.00826 M x 0.02575 L = 0.000212 mol
Moles of NaCl = 0.0183 M x 0.0751 L = 0.001375 mol
Step 4: Determine the limiting reactant.
Since Pb(NO3)2 and NaCl react in a 1:2 ratio (according to the balanced equation), we need twice as many moles of NaCl as Pb(NO3)2. In this case, Pb(NO3)2 is the limiting reactant because we have fewer moles of it.
Step 5: Calculate the concentration of Pb2+ ions in the solution.
After the reaction, all of the Pb(NO3)2 will be converted to PbCl2.
Since Pb(NO3)2 dissociates into one Pb2+ ion per formula unit, the concentration of Pb2+ ions in the solution will equal the moles of Pb(NO3)2 divided by the total volume of the solution.
Concentration of Pb2+ ions = Moles of Pb2+ ions / Total volume of the solution
= 0.000212 mol / (0.02575 L + 0.0751 L) = 0.000212 mol / 0.10085 L ≈ 0.00210 M
Step 6: Compare the concentration of Pb2+ ions with the Ksp.
The Ksp for PbCl2 is 1.7 x 10^-5.
Since the concentration of Pb2+ ions (0.00210 M) is greater than the Ksp (1.7 x 10^-5), a precipitate of PbCl2 will form when lead II nitrate and sodium chloride solutions are mixed.
Therefore, the answer to the question is yes, a precipitate of PbCl2 will form.