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Predicting Precipitates

Demonstrates how solubility constants can be used to predict the formation of precipitates.

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Predicting Precipitates

Barium sulfate is used to obtain clear images of the digestive tract

Credit: Courtesy of Journalist Seaman Apprentice Mike Leporati, US Navy
Source: http://commons.wikimedia.org/wiki/File:US_Navy_060505-N-2832L-010_An_X-Ray_machine_located_aboard_Military_Sealift_Command_%28MSC%29_hospital_ship_USNS_Mercy_%28T-AH_19%29.jpg
License: CC BY-NC 3.0

What do you see?

The invention of the X-ray machine had radically improved medical diagnosis and treatment. For the first time, it was possible to see inside a person’s body to detect broken bones, tumors, obstructions, and other types of problems. Barium sulfate is often used to examine patients with problems of the esophagus, stomach, and intestines. This insoluble compound coats the inside of the tissues and absorbs X-rays, allowing a clear picture of the interior structure of these organs.

Predicting Precipitates

Knowledge of \begin{align*}K_{sp}\end{align*}Ksp values will allow you to be able to predict whether or not a precipitate will form when two solutions are mixed together. For example, suppose that a known solution of barium chloride is mixed with a known solution of sodium sulfate. Barium sulfate is a mostly insoluble compound and so could potentially precipitate from the mixture. However, it is first necessary to calculate the ion product, [Ba2+][SO42−] for the solution. If the value of the ion product is less than the value of the \begin{align*}K_{sp}\end{align*}Ksp, then the solution will remain unsaturated. No precipitate will form because the concentrations are not high enough to begin the precipitation process. If the value of the ion product is greater than the value of the \begin{align*}K_{sp}\end{align*}Ksp, then a precipitate will form. The formation of the precipitate lowers the concentration of each of the ions until the ion product is exactly equal to the \begin{align*}K_{sp}\end{align*}Ksp, at which point precipitation ceases.

Barium sulfate is used as a component of white paint and in certain x-ray imaging processes

Credit: Ondřej Mangl
Source: http://commons.wikimedia.org/wiki/File:S%C3%ADran_barnat%C3%BD.PNG
License: CC BY-NC 3.0

Barium sulfate is used as a component of white pigment for paints and as an agent in certain x-ray imaging processes.[Figure2]

Sample Problem: Predicting Precipitates

Will a precipitate of barium sulfate form when 10.0 mL of 0.0050 M BaCl2 is mixed with 20.0 mL of 0.0020 M Na2SO4?

Step 1: List the known quantities and plan the problem.

Known

  • concentration of BaCl2 = 0.0050 M
  • volume of BaCl2 = 10.0 mL
  • concentration of Na2SO4 = 0.0020 M
  • volume of Na2SO4 = 20.0 mL
  • \begin{align*}K_{sp}\end{align*}Ksp of BaSO4 = 1.1 × 10-10

Unknown

  • ion product [Ba2+][SO42-]
  • if a precipitate forms

The concentration and volume of each solution that is mixed together must be used to calculate the [Ba2+] and the [SO42−]. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the \begin{align*}K_{sp}\end{align*}Ksp to determine if a precipitate forms.

Step 2: Solve.

The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.

\begin{align*}\text{mol Ba}^{2+}&=0.0050 \ \text{M} \times 0.010 \ \text{L}=5.0 \times 10^{-5} \ \text{mol Ba}^{2+} \\ \text{mol SO}_4^{2-}&=0.0020 \ \text{M} \times 0.020 \ \text{L}=4.0 \times 10^{-5} \ \text{mol SO}_4^{2-} \end{align*}mol Ba2+mol SO24=0.0050 M×0.010 L=5.0×105 mol Ba2+=0.0020 M×0.020 L=4.0×105 mol SO24

The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of 0.030 L.

\begin{align*}[\text{Ba}^{2+}] &=\frac{5.0 \times 10^{-5} \ \text{mol}}{0.030 \ \text{L}}=1.7 \times 10^{-3} \ \text{M} \\ \left [\text{SO}_4^{2-}\right ] &=\frac{4.0 \times 10^{-5} \ \text{mol}}{0.030 \ \text{L}}=1.3 \times 10^{-3} \ \text{M} \end{align*}[Ba2+][SO24]=5.0×105 mol0.030 L=1.7×103 M=4.0×105 mol0.030 L=1.3×103 M

Now the ion product is calculated.

\begin{align*}[\text{Ba}^{2+}][\text{SO}_4^{2-}]=(1.7 \times 10^{-3})(1.3 \times 10^{-3})=2.2 \times 10^{-6}\end{align*}

Since the ion product is greater than the \begin{align*}K_{sp}\end{align*}, a precipitate of barium sulfate will form.

Step 3: Think about your result.

Two significant figures are appropriate for the calculated value of the ion product.

 

 

Summary

  • Calculations are shown which allow the prediction of precipitate formation based on \begin{align*}K_{sp}\end{align*}.

Review

  1. What would be the equation for the ion product of BaCl2?
  2. What happens if the ion product is less than the \begin{align*}K_{sp}\end{align*}?
  3. Why did we not need to calculate an ion product for NaCl?

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