A 100.0 mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 mL of 0.200 mL of 0.200 M aqueous magnesium nitrate.
To solve this problem, we will use the concept of stoichiometry and the balanced chemical equation to determine the products formed and the remaining reactants after mixing the two solutions.
The balanced chemical equation for the reaction between potassium hydroxide (KOH) and magnesium nitrate (Mg(NO3)2) is:
2 KOH + Mg(NO3)2 -> 2 KNO3 + Mg(OH)2
Step 1: Determine the limiting reagent
To find the limiting reagent, we need to compare the number of moles of each reactant. The reactant that produces the smallest number of moles of product will be the limiting reagent.
Knowing the initial concentration and volume of each solution, we can calculate the number of moles of KOH and Mg(NO3)2:
moles of KOH = concentration of KOH x volume of KOH = 0.200 M x 0.100 L = 0.0200 moles
moles of Mg(NO3)2 = concentration of Mg(NO3)2 x volume of Mg(NO3)2 = 0.200 M x 0.100 L = 0.0200 moles
Since the ratio of KOH to Mg(NO3)2 in the balanced equation is 2:1, we can see that they have equal moles.
Therefore, neither KOH nor Mg(NO3)2 is limiting. This means that both will react completely.
Step 2: Calculate the number of moles of product formed
From the balanced equation, we can see that for every 2 moles of KOH and 1 mole of Mg(NO3)2, we will get 2 moles of KNO3 and 1 mole of Mg(OH)2.
Since we have equal moles of KOH and Mg(NO3)2 (0.0200 moles each), we can conclude that we will form twice the number of moles of KNO3 and Mg(OH)2.
moles of KNO3 formed = 2 x 0.0200 moles = 0.0400 moles
moles of Mg(OH)2 formed = 0.0200 moles
Step 3: Calculate the concentrations of the products
Given that the final volume is the sum of the initial volumes (100.0 mL + 100.0 mL = 200.0 mL), we can use this to calculate the final concentrations.
Concentration of KNO3 = moles of KNO3 formed / final volume = 0.0400 moles / 0.200 L = 0.200 M
Concentration of Mg(OH)2 = moles of Mg(OH)2 formed / final volume = 0.0200 moles / 0.200 L = 0.100 M
Therefore, the final concentrations of KNO3 and Mg(OH)2 in the resulting solution are 0.200 M and 0.100 M, respectively.
To find out what happens when the two solutions are mixed, we need to consider the reactions that can occur between the potassium hydroxide (KOH) and magnesium nitrate (Mg(NO3)2).
First, let's write out the balanced chemical equation for the reaction between KOH and Mg(NO3)2:
2 KOH + Mg(NO3)2 -> 2 KNO3 + Mg(OH)2
This equation shows that two moles of KOH react with one mole of Mg(NO3)2 to form two moles of KNO3 and one mole of Mg(OH)2.
Given that we have 0.200 M solutions, we can calculate the number of moles in each solution:
For KOH:
0.200 M = 0.200 moles / L
So, in 100.0 mL (0.100 L) aliquot, we have:
0.200 moles / L * 0.100 L = 0.020 moles of KOH
For Mg(NO3)2:
0.200 M = 0.200 moles / L
So, in 100.0 mL (0.100 L) of the solution, we have:
0.200 moles / L * 0.100 L = 0.020 moles of Mg(NO3)2
Comparing the number of moles of KOH and Mg(NO3)2, we see that they have a 1:1 stoichiometric ratio in the balanced equation. This means that all the moles of KOH will react with an equal number of moles of Mg(NO3)2.
Since there is an equal number of moles of both reactants, the reaction goes to completion, and no excess of any reagent remains. All the KOH reacts with Mg(NO3)2, forming KNO3 and Mg(OH)2.
Therefore, when the two solutions are mixed, we obtain a solution containing KNO3 (potassium nitrate) and a precipitate of Mg(OH)2 (magnesium hydroxide).