Обчислення кількості та напрямку випаденого осаду

When I was in college, I studied environmental chemistry. It was super fun! You spend a lot of time hiking outside to collect samples. One of the most fundamental aspects of environmental chemistry is precipitation reactions. Environmental chemistry is very applied. You take what you know and use it in real-world situations. For example, maybe you need to know if something will dissolve in a stream. And if it does, you want to know how much. You can use your knowledge of solubility!

Early on, we learned some qualitative solubility rules. They told us if something was soluble or insoluble. But now, we know equilibrium! Precipitation reactions are just easier equilibrium reactions! Why are they easier? Because precipitates are solids. So, we exclude them from the equilibrium constant.

Now, we can begin to use the solubility product constant. We can use our foundation from equilibrium to solve solubility problems.

Direction of Precipitation Reactions

Just like the equilibrium constant, we can compare $K_{sp}$ to the reaction quotient. The comparison tells us the reaction's direction and if precipitate forms.

• $Q>K_{sp}$

• A compound will precipitate because the solution has excess solute. The precipitate will form until the concentrations are at equilibrium.

• $Q<K_{sp}$

• Nothing precipitates because the solution is not saturated. What happens if we add more solute? It will dissolve until $Q=K_{sp}$

• $Q=K_{sp}$

• The solution is at equilibrium and is saturated. The rates of dissolving and crystallization are equal. So, we'll see some solid formation.

Example Problem

Will a precipitate form when 0.05 liters of $7.0\times10^{-4}$ molar $Pb(NO_{3})_{2}$ is added to 0.45 liters of $7.5\times10^{-5}$ molar $Na_{2}SO_{4}$? The solubility product constant of $PbSO_{4}$ is $6.3\times 10^{-10}$.

• First, we need to use dimensional analysis to calculate the concentrations of lead and sulfate in our solution. Our new solution's volume is $0.05L + 0.45L = 0.50L$.

• $[Pb^{2+}]=0.05L\times7.0\times10^{-4}\frac{mols}{L}\times \frac{1}{0.50L}=7.0\times10^{-5}$

• $[SO_{4}^{2-}]=0.45L\times7.5\times10^{-5}\frac{mols}{L}\times \frac{1}{0.50L}=6.75\times10^{-5}$

• Now, we can calculate the $Q$.

• $Q=[Pb^{2+}][SO_{4}^{2-}]=[7.0\times10^{-5}][6.75\times10^{-5}]=4.73\times10^{-9}$

• $Q>K_{sp}$ so we will see a precipitate.

Calculating Precipitation

We can also do some precipitation calculations. Just like equilibrium calculations, we'll need ICE tables, $K_{sp}$ values, and initial or final concentrations. There is one trick with precipitation. If you want to calculate the precipitate's mass, you need stoichiometry. If you need a refresher, look at the stoichiometry roadmap.

Example Problem

Let's say we mix 500.0 milliliters of 0.25 molar sodium chloride, $NaCl$, and 200.0 milliliters of 0.15 molar silver nitrate, $AgNO_{3}$. The $K_{sp}$ of silver chloride, $AgCl$ is $1.77\times10^{-10}$. What mass of silver chloride will precipitate.

• First, we'll need to write a balanced chemical equation.

• $AgNO_{3} + NaCl \rightleftharpoons AgCl_{(s)} + NaNO_{3(aq)}$

• Compounds with alkali metals are very soluble. So we can assume that sodium nitrate will not precipitate.

• We can write the reaction as a net ionic equation. $Ag^{+} + Cl^{-}\rightleftharpoons AgCl_{(s)}$

• Second, we need to calculate the moles. We can't calculate the concentration of a solid. So, we'll need to work with moles.

• $mols_{Ag}=0.15\frac{mols}{L}\times 0.20L= 0.03mols$

• $mols_{Cl}= 0.25\frac{mols}{L}\times 0.50L= 0.125 mols$

• Next, we need to compare $Q$ to $K_{sp}$. Our total volume is 700.0 milliliters.

• $Q=[Ag^{+}][Cl^{-}]= [\frac{0.03mols}{0.7L}][\frac{.125mols}{0.7L}]= 0.0076.$

• $Q>>K_{sp}$ so we expect all the silver nitrate to precipitate.

• Now, we create our ICE table. The silver ions will be our limiting reagent. The mole ratio is 1:1:1. So, the reactants change will be -0.03mols. The product formed will be 0.03mols. We converted the moles of reactant to moles of product.

$$\begin{array}{|r|c|c|c|} \hline \text{}&Ag^{+}&Cl^{-}&AgCl\\ \hline \text{Initial}&0.03mols&0.125mols&0.0mols\\ \hline \text{Change}&-0.03mols&-0.03mols&+0.03mols\\ \hline \text{Equilibrium}&0.00mols&0.95mols&0.03mols\\ \hline \end{array}$$

• Finally, we can calculate the mass of silver chloride. Its molar mass is $143.3\frac{g}{mol}$.

• $0.03mols\times 143.3\frac{g}{mol}=4.3 g$

(Video) 17.5 Common Ion Effect and Precipitation | General Chemistry

by Chad's Prep

In this section of the video, Chad explains precipitation. You'll learn how to use the solubility product constant and reaction quotient to determine if a precipitate forms. He'll also show you how to calculate the extent of precipitation. He even shows you a way your teacher might try and trick you! After watching this video, you'll understand how to approach any precipitation problem.