# Question 95c1d

Feb 5, 2015

Here's what you're working with. Sodium iodide will fully dissociate in aqueous solution to give

$N a {I}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s N {a}_{\left(a q\right)}^{+} + {I}_{\left(a q\right)}^{-}$

This means that the concentration of ${I}^{-}$ ions will be equal to the concentration of the sodium iodide in the final solution. The moles of $N a I$ added are

$n = C \cdot V = 220 \cdot {10}^{- 3} \text{L" * "0.0255 M" = 5.61 * 10^(-3)"moles}$

The concentration of sodium iodide in the final solution will be

C_("iodide") = n/V_("final") = (5.61 * 10^(-3)"moles")/((220 + 340) * 10^(-3)"L") = "0.01 M"

Therefore, the concentration of ${I}^{-}$ ions will be: $\left[{I}^{-}\right] = \text{0.01 M}$.

Lead (II) chloride however will not fully dissociate in aqueous solution.The solution is unsaturated because the minimum concentration of $P {b}^{2 +}$ ions in solution for a precipitate to form is $\text{0.016 M}$. Since your starting concentration of $P b C {l}_{2}$ is less than that, the lead (II) chloride will dissolve in solution and not remain solid.

As a result, the concentration of the $P {b}^{2 +}$ ions will be equal to that of the lead (II) chloride, or $\left[P {b}^{2 +}\right] = \text{0.005 M}$

SIDE NOTE. You can try and determine that through calculation by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart) and $P b C {l}_{2}$'s solubility product constant, ${K}_{s p} = 1.6 \cdot {10}^{- 5}$.

For the final solution, the concentration of $P {b}^{2 +}$ ions will be equal to

n_("lead (II)") = 340 * 10^(-3)"L" * "0.005 M" = 1.7 * 10^(-3)"moles"

C_("lead (II)") = n/V_("final") = (1.7 * 10^(-3)"moles")/(560 * 10^(-3)"L") = "0.003 M"

This will be the final concentration of $P {b}^{2 +}$ ions in solution: $\left[P {b}^{2 +}\right] = \text{0.003 M}$.

The reaction that is of interest for you will be

$2 {I}_{\left(a q\right)}^{-} + P {b}_{\left(a q\right)}^{2 +} r i g h t \le f t h a r p \infty n s P b {I}_{2 \left(s\right)}$

However, a precipitate will only form if the concentrations of the two ions are large enough. So,

Q_c = [Pb^(2+)] * [I^(-)]^(2) = "0.003 * 0.01^(2) = 3.0 * 10^(-7)

Since this time ${Q}_{c} > {K}_{s p} = 8.3 \cdot {10}^{- 9}$, $P b {I}_{2}$ will form as a precipitate in the final solution.

Now you have to find the limiting reagent in order to determine how much salt is produced. Since the number of ${I}^{-}$ moles is equal to $5.61 \cdot {10}^{- 3}$, and the number of $P {b}^{2 +}$ moles is equal to $1.7 \cdot {10}^{- 3}$, look at the mole ratio you have between the two ions.

One mole of $P {b}^{2 +}$ ions needs 2 moles of ${I}^{-}$ ions $\to$ you have a $1 : 2$ mole ratio between the two. This means that $P {b}^{2 +}$ will be the limiting reagent, since

$2 \cdot 1.7 \cdot {10}^{- 3} = 3.4 \cdot {10}^{- 3} < 5.61 \cdot {10}^{- 3}$

This means that the number of moles of $P b {I}_{2}$ produced will be

$1.7 \cdot {10}^{- 3} \text{moles" * ("1 mole Pb"^(2+))/("1 mole PbI"_2) = 1.7 * 10^(-3)"moles}$

Therefore,

$1.7 \cdot {10}^{- 3} \text{moles" * ("461 g")/("1 mole") = "0.78 g}$ $P b {I}_{2}$

Now for the ions that remain in solution. Since $P {b}^{2 +}$ was the limiting reagent, it was fully consumed in the reaction; this means that its remaining concentration in solution will be zero. The concentration of the ${I}^{-}$ ions is determined by the excess moles

n_("excess"I^(-)) = 5.6 * 10^(-3) - 2 * 1.7 * 10^(-3) = 2.2 * 10^(-3)"moles"

Thus, the remaining concentration of ${I}^{-}$ ions will be

[I^(-)] = (2.2 * 10^(-3)"moles")/(560 * 10^(-3)"L") = "0.0040 M"#