Chapter 15, Problem 15.2KSP

Determine the concentrations of Pb²⁺ and I⁻ in a saturated solution of PbI₂ (K_sp for PbI₂ is 1.4 × 10⁻⁸.)

(a) 0.0015 M and 0.0030 M

(b) 0.0015 M and 0.0015 M

(c) 0.0015 M and 0.00075 M

(d) 0.0019 M and 0.0019 M

(e) 0.0019 M and 0.0038 M

Interpretation Introduction

Interpretation:

The concentrations values for the given ions should be determined using the given solubility product value.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If K value is less than one then the respective reaction will move to the left side and if K value is higher (or) greater than one then respective reaction will move to the right side of reaction.

Solubility product: It is used to express the concentration of dissolved ions whose stoichiometric coefficients are raised to power of it. It expresses the equilibrium between solids and its respective ions in the solution.

Answer to Problem 15.2KSP

The concentration values for ${\text{Pb}}^{\text{2+}}$ and ${\text{I}}^{\text{-}}$ is $\left(\text{a}\right)0.0015M\text{and}0.0030M$.

Explanation of Solution

Reason for correct option:

The general chemical equation that represents the given conditions is as follows,

${\text{PbI}}_{\text{2}}{}_{\left(s\right)}\stackrel{}{\to }{\text{Pb}}^{\text{2+}}{}_{\left(aq\right)}\text{+}2{\text{I}}^{\text{-}}{}_{\left(aq\right)}$

Given,

The solubility product for ${\text{PbI}}_{\text{2}}$ is $\text{1}{\text{.4×10}}^{\text{-8}}$.

Considering the given reaction we set up (ICE) equilibrium table as follows,

$\begin{array}{l}{\text{PbI}}_{\text{2}}{}_{\left(s\right)}\stackrel{}{\to }{\text{Pb}}^{\text{2+}}{}_{\left(aq\right)}\text{+}2{\text{I}}^{\text{-}}{}_{\left(aq\right)}\\ \text{Initial concentration: }x00\\ \text{After some time : }-x+x+2x\\ -----------------------------\\ \text{At the end of the reaction: }x-xx2x\end{array}$

The concentration for ${\text{Pb}}^{\text{2+}}$ and ${\text{I}}^{\text{-}}$ is calculated as follows using the given solubility product value.

$\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{I}}^{\text{-}}\right]}^{\text{2}}\\ \text{1}{\text{.4×10}}^{\text{-8}}\text{=}\left[\text{x}\right]\left[{\left(\text{2x}\right)}^{\text{2}}\right]\\ 4{\text{x}}^{\text{3}}=\text{1}{\text{.4×10}}^{\text{-8}}\\ \end{array}$

$\begin{array}{l}{\text{x}}^{\text{3}}=\frac{\text{1}{\text{.4×10}}^{\text{-8}}}{4}\\ {\text{x}}^{\text{3}}=3.5\times {10}^{-9}\\ \text{x}\text{=}\text{1}\text{.52}\times {10}^{-3}\end{array}$

$\begin{array}{l}{\text{The concentration for Pb}}^{\text{2+}}\text{,x}\text{=}\text{1}{\text{.52×10}}^{\text{-3}}\\ =0.0015M\\ {\text{The concentration for I}}^{\text{-}}\text{,}\text{2x}\text{=}\text{2×1}{\text{.52×10}}^{\text{-3}}\text{=}\text{3}{\text{.04×10}}^{\text{-3}}\\ =0.0030M\end{array}$

Therefore, the value for x was determined and the concentration of ${\text{Pb}}^{\text{2+}}$ is equal to $0.0015M$ and the value for ${\text{I}}^{\text{-}}$ is $0.0030M$ since its final concentration is equal to $\text{2x}$.

Hence, the correct option is option (a).

Reason for in-correct options:

Consider all options other than the correct one.

$\begin{array}{l}\left(\text{b}\right)\text{0}\text{.0015M}\text{and}\text{0}\text{.0015M}\\ \left(\text{c}\right)\text{0}\text{.0015M}\text{and}\text{0}\text{.00075M}\\ \left(\text{d}\right)\text{0}\text{.0019M}\text{and}\text{0}\text{.0019M}\\ \left(\text{e}\right)\text{0}\text{.0019M}\text{and}\text{0}\text{.0038M}\end{array}$

These options show us that they do not contain the correct value which is obtained from the calculation using solubility product formula and the ice table for the given chemical reaction.

Hence, the incorrect options are option (b), (c) (d)and (e).

Conclusion

The correct option that contains the concentration values for the given reaction was determined by using solubility product formula.