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Determination of molar solubility from K sp

The K sp of copper(I) bromide, CuBr, is 6.3 × 10 –9 . Calculate the molar solubility of copper bromide.

Solution

The solubility product constant of copper(I) bromide is 6.3 × 10 –9 .

The reaction is:

CuBr ( s ) Cu + ( a q ) + Br ( a q )

First, write out the solubility product equilibrium constant expression:

K sp = [ Cu + ] [ Br ]

Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the K sp :

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following 0, x, 0 plus x equals x.

At equilibrium:

K sp = [ Cu + ] [ Br ]
6.3 × 10 9 = ( x ) ( x ) = x 2
x = ( 6.3 × 10 9 ) = 7.9 × 10 5

Therefore, the molar solubility of CuBr is 7.9 × 10 –5 M .

Check your learning

The K sp of AgI is 1.5 × 10 –16 . Calculate the molar solubility of silver iodide.

Answer:

1.2 × 10 –8 M

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Determination of molar solubility from K sp , part ii

The K sp of calcium hydroxide, Ca(OH) 2 , is 1.3 × 10 –6 . Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product constant of calcium hydroxide is 1.3 × 10 –6 .

The reaction is:

Ca(OH) 2 ( s ) Ca 2+ ( a q ) + 2OH ( a q )

First, write out the solubility product equilibrium constant expression:

K sp = [ Ca 2+ ] [ OH ] 2

Create an ICE table, leaving the Ca(OH) 2 column empty as it is a solid and does not contribute to the K sp :

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, and 0 plus x equals x. The third column has the following 0, 2 x, and 0 plus 2 x equals 2 x.

At equilibrium:

K sp = [ Ca 2+ ] [ OH ] 2
1.3 × 10 6 = ( x ) ( 2 x ) 2 = ( x ) ( 4 x 2 ) = 4 x 3
x = 1.3 × 10 6 4 3 = 6.9 × 10 3

Therefore, the molar solubility of Ca(OH) 2 is 1.3 × 10 –2 M .

Check your learning

The K sp of PbI 2 is 1.4 × 10 –8 . Calculate the molar solubility of lead(II) iodide.

Answer:

1.5 × 10 –3 M

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Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. [link] shows how to perform those unit conversions before determining the solubility product equilibrium.

Determination of K sp From gram solubility

Many of the pigments used by artists in oil-based paints ( [link] ) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO 4 , is 4.6 × 10 –6 g/L. Determine the solubility product equilibrium constant for PbCrO 4 .

A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface.
Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO 4 ), examples include Prussian blue (Fe 7 (CN) 18 ), the reddish-orange color vermilion (HgS), and green color veridian (Cr 2 O 3 ). (credit: Sonny Abesamis)

Solution

We are given the solubility of PbCrO 4 in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb 2+ and CrO 4 2− , then K sp :

This figure shows four horizontally oriented rectangles. The first three from the left are shaded green and the last one at the right is shaded white. Right pointing arrows between the rectangles are labeled “1,” “2,” and “3” moving left to right across the diagram. The first rectangle is labeled “Solubility of P b C r O subscript 4, in g divdided by L.” The second rectangle is labeled “[ P b C r O subscript 4 ], in m o l divided by L.” The third is labeled “[ P b superscript 2 plus] and [ C r O subscript 4 superscript 2 negative ].” The fourth rectangle is labeled “K subscript s p.”


  1. Use the molar mass of PbCrO 4 ( 323.2 g 1 mol ) to convert the solubility of PbCrO 4 in grams per liter into moles per liter:

    [ PbCrO 4 ] = 4.6 × 10 6 g PbCrO 4 1 L × 1 mol PbCrO 4 323.2 g PbCrO 4 = 1.4 × 10 8 mol PbCrO 4 1 L = 1.4 × 10 8 M
  2. The chemical equation for the dissolution indicates that 1 mol of PbCrO 4 gives 1 mol of Pb 2+ (aq) and 1 mol of CrO 4 2− ( a q ) :

    PbCrO 4 ( s ) Pb 2+ ( a q ) + CrO 4 2− ( a q )

    Thus, both [Pb 2+ ] and [ CrO 4 2− ] are equal to the molar solubility of PbCrO 4 :

    [ Pb 2+ ] = [ CrO 4 2− ] = 1.4 × 10 8 M
  3. Solve. K sp = [Pb 2+ ] [ CrO 4 2− ] = (1.4 × 10 –8 )(1.4 × 10 –8 ) = 2.0 × 10 –16

Check your learning

The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.46 grams per liter at 20 °C. What is its solubility product?

Answer:

1.7 × 10 –4

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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