Question

A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

Short answer

a. The balanced chemical equation for the reaction is: \(KOH(aq) + Mg(NO_3)_2(aq) \longrightarrow Mg(OH)_2(s) + 2KNO_3(aq)\) b. The precipitate formed is magnesium hydroxide \((Mg(OH)_2)\). c. The mass of the precipitate produced is \(1.166 g\). d. The concentrations of the remaining ions in the solution are: \(K^+\): \(0.200 M\), \(NO_3^-\): \(0.200 M\).

Step-by-step solution

a. Balanced Chemical Equation

To determine the balanced chemical equation, first, write down the formulas of the two reactants and then predict the products through double displacement. Potassium hydroxide: \(KOH\), aqueous Magnesium nitrate: \(Mg(NO_3)_2\), aqueous Double displacement: Potassium will swap places with magnesium in each compound. \(KOH(aq) + Mg(NO_3)_2(aq) \longrightarrow Mg(OH)_2(s) + 2KNO_3(aq)\) Now we have a balanced chemical equation for the reaction taking place.

b. Precipitate Formation

In the balanced chemical equation, we can see that magnesium hydroxide \(( Mg(OH)_2 )\) is one of the products, and it is in the solid phase. Therefore, magnesium hydroxide forms a precipitate.

c. Mass of Precipitate Produced

First, we need to determine the limiting reactant from the given concentrations of the two reactants: Concentration of \(KOH\) = \(0.200 M\) Volume of \(KOH\) = \(100.0 mL\) Concentration of \(Mg(NO_3)_2\) = \(0.200 M\) Volume of \(Mg(NO_3)_2\) = \(100.0 mL\) Convert the volumes to liters: Volume of \(KOH\) = \(0.100 L\) Volume of \(Mg(NO_3)_2\) = \(0.100 L\) Calculate the moles of each reactant: moles of \(KOH = 0.200 M \times 0.100 L = 0.020\; mol\) moles of \(Mg(NO_3)_2 = 0.200 M \times 0.100 L = 0.020\; mol\) Using the stoichiometry of the balanced chemical equation, we can calculate the moles of \(Mg(OH)_2\) formed: \(1 \;mol\; KOH \rightarrow 1 \;mol\; Mg(OH)_2\) moles of \(Mg(OH)_2 = 0.020\; mol\) Now, we can convert moles to grams using molar mass: Molar mass of \(Mg(OH)_2 = 24.3+2(16.0+1.0) = 58.3 g/mol\) Mass of precipitate produced = moles of \(Mg(OH)_2 * Molar mass of Mg(OH)_2\) Mass of precipitate produced = \(0.020\; mol \times 58.3 g/mol = 1.166 g\)

d. Concentration of Ions Remaining in Solution

Since all the moles of \(KOH\) reacted with all the moles of \(Mg(NO_3)_2\) and formed a precipitate, there is no excess of any reactant. The remaining ions are those of the soluble salt \(KNO_3\). The balanced equation shows that two moles of \(KNO_3\) form for every mole of \(KOH\), so: moles of \(K^+\) = moles of \(NO_3^-\) = \(2 \times moles\; of\; KOH = 2 \times 0.020\; mol = 0.040\; mol\) Since the volume of the reaction mixture is the sum of the initial volumes: Final volume = \(100.0 mL + 100.0 mL = 200.0 mL = 0.200 L\) Now, we can calculate the concentration of each ion remaining in the solution: Concentration of \(K^+\) = moles of \(K^+\) / total volume Concentration of \(K^+\) = \(0.040\; mol / 0.200 L = 0.200 M\) Concentration of \(NO_3^-\) = moles of \(NO_3^-\) / total volume Concentration of \(NO_3^-\) = \(0.040\; mol / 0.200 L = 0.200 M\) Thus, the concentrations of the remaining ions in the solution are: \(K^+\): \(0.200 M\) \(NO_3^-\): \(0.200 M\)

Key concepts

Precipitation

Precipitation in chemical reactions occurs when two soluble substances react to form an insoluble solid, known as the precipitate. In our example, when potassium hydroxide (KOH) and magnesium nitrate ( Mg(NO_3)_2 ) are mixed, a precipitation reaction happens. The magnesium ( Mg^{2+} ) cations from the magnesium nitrate combine with the hydroxide ( OH^- ) anions from the potassium hydroxide to form magnesium hydroxide ( Mg(OH)_2 ).

• Magnesium hydroxide is insoluble in water, causing it to form a solid—a precipitate.
• This solid separates from the liquid, settling at the bottom of the container.

This forms the crux of precipitation reactions; soluble ions come together to form an insoluble compound.

Stoichiometry

Stoichiometry helps to determine the quantitative relationships in chemical reactions. This means calculating the amounts of reactants and products. In our problem, we used stoichiometry to identify the limiting reactant and the amount of precipitate formed. Let's break it down:

• First, convert the volume of each reactant from milliliters to liters.


• Then, use the molarity to find the number of moles for each reactant: 0.200 M KOH and 0.200 M Mg(NO_3)_2 , each in 0.100 L.
• By comparing the mole ratio from the balanced equation, verify the stoichiometry: 1 mole of KOH reacts with 1 mole of Mg(NO_3)_2 to produce 1 mole of Mg(OH)_2 .

This calculation helped us determine that 0.020 moles of each reactant are used, producing 0.020 moles of magnesium hydroxide. Thus, stoichiometry guided us to calculate the precipitate mass.

Ionic Equations

Ionic equations showcase the ions that participate in a chemical reaction, clearly illustrating reactants and products in their ionic forms. The initial molecular equation for our reaction is:\[KOH (aq) + Mg(NO_3)_2 (aq) \rightarrow Mg(OH)_2 (s) + 2 KNO_3 (aq)\]This can be broken down into the ionic equation, which displays the ions in aqueous solutions:

• Potassium ions (K^+) and hydroxide ions (OH^-) from KOH.


• Magnesium ions (Mg^{2+}) and nitrate ions (NO_3^-) from Mg(NO_3)_2.

Since nitrates and potassium salts are soluble, they do not form a precipitate.

• The net ionic equation for the reaction shows only the ions that contribute to the precipitate formation:

\[Mg^{2+}(aq) + 2 OH^-(aq) \rightarrow Mg(OH)_2(s)\]This simplified form excludes spectator ions, focusing only on the formation of Mg(OH)_2.

Solution Concentration

Solution concentration indicates how much solute is present in a given volume of solvent, usually expressed in molarity ( M ). After the reaction, we calculated the concentration of ions remaining in solution:

• The total volume after mixing is 200.0 mL or 0.200 L.


• Once the precipitation of magnesium hydroxide occurs, only the ions from the soluble potassium nitrate remain in the solution: K^+ and NO_3^- ions.

With a 1:2 ratio in the balanced equation, there are 0.040 moles each of K^+ and NO_3^- ions, divided by the total volume of the solution:

• Concentration of K^+ = 0.040 ext{ mol} / 0.200 ext{ L} = 0.200 M .
• Concentration of NO_3^- = 0.040 ext{ mol} / 0.200 ext{ L} = 0.200 M .

This calculation shows that understanding concentrations is key to knowing the remaining make-up of the solution.