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18.6: Precipitation

Last updated

Jul 16, 2022
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- 18.5: Solubility Equilibria and the Solubility Product Constant
- 18.7: Complex Ion Equilibria

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Learning Objectives

Calculate ion concentrations to maintain a heterogeneous equilibrium.
Calculate pH required to precipitate a metal hydroxide.
Design experiments to separate metal ions in a solution of mixtures of metals.

A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$ , $\ce{Br-}$ , $\ce{SO4^2-}$ , $\ce{CO3^2-}$ , $\ce{S^2-}$ , $\ce{Cr2O4^2-}$ , $\ce{PO4^2-}$ , $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.

What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.

All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$ , $\ce{ClO3-}$ , $\ce{NO2-}$ , $\ce{HCOO-}$ , and $\ce{CH3COO-}$ .
All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$ , $\ce{Hg2^2+}$ , and $\ce{Pb^2+}$ . $\ce{CaF2}$ , $\ce{BaF2}$ , and $\ce{PbF2}$ are also insoluble.
All sulfates are soluble, except those of $\ce{Ba^2+}$ , $\ce{Sr^2+}$ , and $\ce{Pb^2+}$ . The doubly charged sulfates are usually less soluble than halides and nitrates.
Most singly charge cations $\ce{K+}$ , $\ce{Na+}$ , $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.

These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.

Chemical Separation of Metal Ions

Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The K_sp values for various salts are valuable information, and some data are given in Table E3. In the first two examples, we show how barium and strontium can be separated as chromate.

Example $\PageIndex{1}$

The K_sp for strontium chromate is $3.6 \times 10^{-5}$ and the K_sp for barium chromate is $1.2 \times10^{-10}$ . What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?

Solution
Since the K_sp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Q_sp for $\ce{BaCrO4}$ also increases from zero to K_sp of $\ce{BaCrO4}$ , at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Q_sp for $\ce{SrCrO4}$ increases to K_sp of $\ce{SrCrO4}$ ; it then precipitates.

Let us write the equilibrium equations and data down to help us think. Let $x$ be the concentration of chromate to precipitate $\ce{Sr^2+}$ , and $y$ be that to precipitate $\ce{Ba^2+}$ :

$\ce{SrCrO4(s) \rightarrow Sr^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$

According to the definition of K_sp we have we have $K_{\ce{sp}} = (0.30)(x) = 3.6 \times 10^{-5}$ . Solving for $x$ gives

$x = \dfrac{3.6 \times 10^{-5}}{0.30} = 1.2 \times 10^{-4} M \nonumber$

Further, let $y$ be the concentration of chromate to precipitate precipitate $\ce{Ba^2+}$ :

$\ce{BaCrO4(s) \rightarrow Ba^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$

with $K_{\ce{sp}} = (0.30)(y) = 1.2 \times 10^{-10}$ . Solving for $y$ gives

$y = \dfrac{1.2 \times 10^{-10}}{0.30} = 4.0 \times 10^{-10} \;M \nonumber$

The K_sp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained less than $1.2 \times 10^{-4} M$ , then all $\ce{Sr^2+}$ ions will remain in the solution.

Discussion

In reality, controling the increase of $\ce{[CrO4^2- ]}$ is very difficult.

Example $\PageIndex{2}$

The K_sp for strontium chromate is $3.6\times 10^{-5}$ and the K_sp for barium chromate is $1.2\times 10^{-10}$ . Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$ . Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.

Solution

From the solution given in Example $\PageIndex{1}$ , $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated at $3.6 \times 10^{-4} = 1.2\times 10^{-10}$ .

The K_sp of $\ce{BaCrO4}$ .

Thus,

$\ce{[Ba^2+]} = 3.33 \times 10^{-7}\, M \nonumber$

Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$ , the molar ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is

$\dfrac{0.30}{3.33 \times 10^{-7}} = 9.0 \times 10^{5}. \nonumber$

Hence, the amount of $\ce{Ba^2+}$ ion in the solid is only $1 \times 10^{-6}$ (i.e., 1 ppm) of all metal ions, providing that all the solid was removed when

$\ce{[CrO4^{2-}]} = 3.6 \times 10^{-4} M. \nonumber$

Discussion

The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.

Example $\PageIndex{3}$

What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?

Solution

The K_sp's for salts of silver and lead are required. We list the K_sp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts K_sp.

Solutions to Example 17.6.3
Salt	K_sp	Salt	K_sp
$\ce{AgCl}$	$1.8 \times 10^{-10}$	$\ce{Ag2SO4}$	$1.4\times 10^{-5}$
$\ce{Hg2Cl2}$	$1.3\times 10^{-18}$	$\ce{BaSO4}$	$1.1\times 10^{-10}$
$\ce{PbCl2}$	$1.7 \times 10^{-5}$	$\ce{CaSO4}$	$2.4\times 10^{-5}$
		$\ce{PbSO4}$	$6.3\times 10^{-7}$
		$\ce{SrSO4}$	$3.2\times 10^{-7}$

Because the K_sp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$ , apppears a good choice of negative ions for their separation.

The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80^oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.

Discussion

Find more detailed information about the solubility of lead chloride as a function of temperature.

Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$ ?

The Separation of Two Ions by a Difference in Solubility: The Separation of Two Ions by a Difference in Solubility(opens in new window) [youtu.be]

Learning Objectives

Chemical Separation of Metal Ions

Example 18.6.1\PageIndex{1}

Discussion

Example 18.6.2\PageIndex{2}

Solution

Discussion

Example 18.6.3\PageIndex{3}

Solution

Discussion

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Example $\PageIndex{1}$

Example $\PageIndex{2}$

Example $\PageIndex{3}$