When 10.0 mL of 0.012 M pb(NO3)2 is mixed with 10.0 mL of 0.030 M KI, a yellow precipitate of PbI2(s) forms.
a)Calculate the molarity of [Pb^2+]
b)Calculate the initial molarity of [I^-]
c)On measuring the equilibrium concentrations of [I^-] it came out to be 8.0 x 10^-3 M. Calculate the molarity of [I^-] precipitated.
The problem looks simple enough; however, after reading and re-reading it I don't understand it at all. Do you know Ksp? Do you want the concn Pb^+2 before mixing, after mixing but before pptn of PbI2? For part b, molarity of I^- before mixing or after mixing and before or after pptn of PbI2.
For part c, if M = moles/L of soln and PbI2 is a ppt (and insoluble), how can it make sense to calculate a M?
this is what the question exactly says
Do you know Ksp?
no it is not given
The best I can do is leave this for another tutor. Perhaps another tutor can make sense of it. Personally, I think the problems needs to be clarified.
GOOD TIMES
GOOD TIMES
SIGH
To answer these questions, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction that occurs between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI):
Pb(NO3)2 + 2KI → PbI2(s) + 2KNO3
a) To calculate the molarity of [Pb²⁺], we first need to determine the moles of Pb²⁺ ions present in the solution. We know that 10.0 mL of 0.012 M Pb(NO3)2 is mixed, so:
moles of Pb²⁺ = volume (in L) × molarity
= 10.0 mL × (1 L / 1000 mL) × 0.012 M
= 0.00012 moles
Since the stoichiometry of the balanced equation is 1:1 for Pb(NO3)2 to Pb²⁺ ions, the molarity of [Pb²⁺] is the same as the moles:
[Pb²⁺] = moles / volume (in L)
= 0.00012 moles / 0.020 L
= 0.006 M
b) To calculate the initial molarity of [I⁻], we use a similar process. We know that also 10.0 mL of 0.030 M KI is mixed:
moles of I⁻ = volume (in L) × molarity
= 10.0 mL × (1 L / 1000 mL) × 0.030 M
= 0.0003 moles
However, the stoichiometry of the balanced equation is 2:1 for KI to I⁻ ions. So, we need to divide the moles by 2:
initial molarity of [I⁻] = moles / volume (in L)
= 0.0003 moles / 0.020 L
= 0.015 M
c) On measuring the equilibrium concentrations of [I⁻], it is found to be 8.0 x 10⁻³ M. To calculate the molarity of [I⁻] precipitated, we need to find the change in concentration of [I⁻] from the initial concentration to the equilibrium concentration:
Change in [I⁻] = Initial [I⁻] - Equilibrium [I⁻]
= 0.015 M - 8.0 x 10⁻³ M
= 0.015 M - 0.008 M
= 0.007 M
Therefore, the molarity of [I⁻] precipitated is 0.007 M.