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.
sorry didn't mean to respost this
Answered below. I was working on the response when you posted a repeat. Give us a little time to respond. It takes time to type this material on a board that was never meant for chemistry, exponents, subscripts, etc.
To determine whether a precipitate will form when two solutions are mixed, you need to compare the calculated ion product, Q, with the solubility product constant, Ksp. If Q is greater than Ksp, a precipitate will form.
Here's how you can set up the calculation to find out if a precipitate of PbCl2 will form:
Step 1: Write the balanced chemical equation for the reaction between lead II nitrate (Pb(NO3)2) and sodium chloride (NaCl):
Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)
Step 2: Calculate the moles of lead II nitrate (Pb(NO3)2) and sodium chloride (NaCl) individually.
moles of Pb(NO3)2 = volume (L) × concentration (M) = 0.02575 L × 0.00826 M = 0.000212675 mol
moles of NaCl = volume (L) × concentration (M) = 0.0751 L × 0.0183 M = 0.00137613 mol
Step 3: Determine the limiting reagent by comparing the moles of each reactant. The reactant that produces fewer moles of the precipitate, PbCl2, is the limiting reagent.
From the balanced equation, the stoichiometry between Pb(NO3)2 and PbCl2 is 1:1. Therefore, the moles of PbCl2 formed will be equal to the moles of Pb(NO3)2.
Since both reactants have a 1:1 stoichiometric ratio with PbCl2, the limiting reagent is Pb(NO3)2.
Step 4: Calculate the moles of PbCl2 that will form based on Pb(NO3)2 (the limiting reagent).
moles of PbCl2 = moles of Pb(NO3)2 = 0.000212675 mol
Step 5: Calculate the concentration of Pb2+ and Cl- ions in the final solution.
concentration of Pb2+ ions = moles of Pb2+ / total volume of solution (L)
= 0.000212675 mol / (0.02575 L + 0.0751 L) = 0.00221876 M
concentration of Cl- ions = 2 × moles of PbCl2 / total volume of solution (L)
= 2 × 0.000212675 mol / (0.02575 L + 0.0751 L) = 0.00443752 M
Step 6: Calculate the ion product, Q, by multiplying the concentrations of Pb2+ and Cl- ions.
Q = [Pb2+][Cl-] = 0.00221876 M × 0.00443752 M = 9.842 × 10^-6
Step 7: Finally, compare the calculated ion product, Q, with the solubility product constant, Ksp.
If Q > Ksp (1.7 × 10^-5), then a precipitate of PbCl2 will form.
In this case, Q (9.842 × 10^-6) is less than Ksp, so a precipitate of PbCl2 will not form when the two solutions are mixed.
By following these steps, you can determine whether a precipitate will form by comparing Q and Ksp values.