General chemistry question

Question

. A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 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

I need help with d, thanks.

Answer 1

Hey Nelly! Thank you so much for asking this complex general chemistry question.

Hopefully the following response helps you get on the right track, at least a little bit. Let me know if you have any follow up questions.

When calculating the concentration of each ion remaining in solution after full and complete precipitation of Mg(OH)2, or magnesium hydroxide, it is imperative to:

first (1) balance the chemical equation/reaction which can be written as follows: 2 KOH + Mg(NO3)2 -> 2 KNO3 + Mg(OH)2

Second, (2): confirm which species is the precipitate that is formed. In this case, it is magnesium hydroxide, or Mg(OH)2.

(3): Determine the limiting reagant in the overall reaction to form the precipitate, which will then allow us to determine the mass of that precipitate formed.

d. Calculate the concentration of each ion remaining in solution after precipitation is complete

Number of moles of KOH:

Molarity of KOH (M1) = 0.200 M

Volume of KOH (V1) = 100.0 mL = 0.100 L

Number of moles of KOH = M1 * V1

Number of moles of KOH = 0.200 M * 0.100 L = 0.020 moles

Number of moles of Mg(NO3)2:

Molarity of Mg(NO3)2 (M2) = 0.200 M

Volume of Mg(NO3)2 (V2) = 100.0 mL = 0.100 L

Number of moles of Mg(NO3)2 = M2 * V2

Number of moles of Mg(NO3)2 = 0.200 M * 0.100 L = 0.020 moles

Since the mole ratio between KOH and Mg(NO3)2 is 2:1 (from the balanced chemical equation), both reactants are present in equal moles, and therefore, Mg(NO3)2 is the limiting reactant.

(4): calculate the moles of Mg(OH)2 precipitate formed. From the balanced chemical equation, we know that 1 mole of Mg(NO3)2 produces 1 mole of Mg(OH)2.

Number of moles of Mg(OH)2 = Number of moles of limiting reactant (Mg(NO3)2) = 0.020 moles

Molar mass of Mg(OH)2 = 58.33 g/mol (magnesium: 24.31 g/mol, oxygen: 16.00 g/mol, hydrogen: 2.02 g/mol)

Mass of precipitate (Mg(OH)2) = Number of moles of Mg(OH)2 * Molar mass of Mg(OH)2

Mass of precipitate (Mg(OH)2) = 0.020 moles * 58.33 g/mol ≈ 1.166 g

(5) Calculate the Concentration of Each Ion Remaining in Solution:

After precipitation is complete, the Mg2+ ions have reacted to form the precipitate, leaving potassium ions (K+) and nitrate ions (NO3-) in solution.

Since potassium nitrate (KNO3) is soluble in water, it dissociates completely, and we have 2 moles of K+ ions and 2 moles of NO3- ions for every mole of KNO3.

The final volume of the solution is the sum of the initial volumes of KOH and Mg(NO3)2 solutions, which is 100.0 mL + 100.0 mL = 200.0 mL = 0.200 L.

Concentration of K+ ions:

Number of moles of K+ ions = 2 moles (from 1 mole of KNO3)

Concentration of K+ ions = Number of moles of K+ ions / Final volume of the solution

Concentration of K+ ions = 2 moles / 0.200 L = 10.0 M

Concentration of NO3- ions:

Number of moles of NO3- ions = 2 moles (from 1 mole of KNO3)

Concentration of NO3- ions = Number of moles of NO3- ions / Final volume of the solution

Concentration of NO3- ions = 2 moles / 0.200 L = 10.0 M

So, the concentration of K+ ions and NO3- ions remaining in solution after precipitation is complete is 10.0 M for both.