Neutralization & Precipitation

Predicting Precipitation Formation

Solubility curves are used in chemical applications to determine whether a reaction will result in the formation of a precipitate. For example,

$\mathrm{Ag}{}^{+}+\mathrm{Cl}{}^{-}\rightleftharpoons \mathrm{AgCl}{}_{(\mathrm{s})}$

$[A{g}^{+}]$

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$[C{l}^{-}]$

This curve gives all the possible combinations of concentrations which are likely to result in the formation of a precipitate at a given temperature. If the product of concentrations is above the curve, a precipitate forms. A product below the curve indicates that the solution is unsaturated while products on the curve result in saturated solutions.

Example 1

50.0 mL of both 0.100 mol/L $\mathrm{Pb}(\mathrm{NO}{}_3){}_2$ and 0.015 mol/L KI solutions are mixed together. A yellow precipitate forms. Determine the type of precipitate formed in a quantitative manner.

Solution: $\mathrm{Pb}(\mathrm{NO}{}_3){}_2{}_{(\mathrm{s})}\to \mathrm{Pb}{}^{2+}{}_{(\mathrm{aq})}+2\mathrm{NO}{}_3{}^{-}{}_{(\mathrm{aq})}$

After examining solubility rules, $\mathrm{Pb}(\mathrm{NO}{}_3){}_2{}_{(\mathrm{s})}$ is likely to dissociate completely into ions.

Amount of $\mathrm{Pb}(\mathrm{NO}{}_3){}_2$ = C × V = 0.100 mol/L × 0.050 L = 0.005 mol

Note: $[\mathrm{Pb}{}^{2+}{]}_{new}$ = 0.005 mol (after mixing) 0.100 L = 0.05 mol/L

$[\mathrm{NO}{}_3{}^{-}{]}_{new}$ = 2(0.005 mol) 0.100 L = 0.10 mol/L

$\mathrm{KI}{}_{(\mathrm{s})}\to \mathrm{K}{}^{+}{}_{(\mathrm{aq})}+\mathrm{I}{}^{-}{}_{(\mathrm{aq})}$

Again, KI is expected to completely dissociate into ions as predicted by solubility rules.

Amount of KI = C × V = 0.015 mol/L × 0.050 L = 0.000 75 mol

$[K^{+}{]}_{new}$ = $[I^{-}{]}_{new}$ = 0.000 75 mol (after mixing) 0.100 L = 7.5 × 10^-3 mol/L

Possible Precipitates

• According to solubility rules, $\mathrm{KNO}{}_3$ is not likely to form.
• $\mathrm{PbI}{}_2$ is a possible precipitate resulting the following equilibrium.

$\mathrm{PbI}{}_2{}_{(\mathrm{s})}\rightleftharpoons \mathrm{Pb}{}^{2+}{}_{(\mathrm{aq})}+2\mathrm{I}{}^{-}{}_{(\mathrm{aq})}$

$K_{sp}$ = $[\mathrm{Pb}{}^{2+}][\mathrm{I}{}^{-}{]}^2$ $K_{sp}$ = (0.05 mol/L)(7.5 × 10^-3 mol/L)^2 = 2.8 × 10^-6 mol^3/L

However, $K_{sp}$ = 8.5 × 10^-9 mol^3/L^3 (from tables)

Since the experimental product of $[\mathrm{Pb}{}^{2+}]$ and $[\mathrm{I}{}^{-}{]}^2$ is greater than the $K_{sp}$ (2.8 × 10^-6 > 8.5 × 10^-9), a precipitate will form.

If the product of $[\mathrm{Pb}{}^{2+}]$ and $[\mathrm{I}{}^{-}{]}^2$ was less than $K_{sp}$, no precipitate would form.

Example 2

Should precipitation occur when 50.0 mL of 5.00 × 10^-4 mol/L calcium nitrate solution is mixed with 50.0 mL of 2.00 × 10^-4 mol/L sodium fluoride solution. The $K_{sp}$ for calcium fluoride is 1.7 × 10^-10 (mol/L)^3.

Solution:

• $\mathrm{NaNO}{}_3$ is not likely to precipitate according to solubility rules. $\mathrm{CaF}{}_2{}_{(\mathrm{s})}\rightleftharpoons \mathrm{Ca}{}^{2+}{}_{(\mathrm{aq})}+2\mathrm{F}{}^{-}{}_{(\mathrm{aq})}$

After mixing, $[\mathrm{Ca}{}^{2+}]$ = 50 mL × 5.0 × 10^-4 mol/L 100 mL = 2.5 × 10^-4 mol/L

and $[\mathrm{F}{}^{-}]$ = 50 mL × 2.0 × 10^-4 mol/L 100 mL = 1.0 × 10^-4 mol/L

Since $K_{sp}$ = $[\mathrm{Ca}{}^{2+}][\mathrm{F}{}^{-}{]}^2$

Then $K_{sp}$ = (2.5 × 10^-4 mol/L)(1.0 × 10^-4 mol/L)^2 = 2.5 × 10^-12 (mol/L)

Since the experimentally determined solubility product [2.5 × 10^-12 (mol/L)^3] is less than the theoretical $K_{sp}$ [1.7 × 10^-10 (mol/L)^3], a precipitate will not form.

Assignment: P. 468 # 1-4