The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.

Step by step solution

01

Understand the initial conditions

Solution A contains copper(II) nitrate with a volume of 2.00 L and a concentration of 2.00 M. Solution B contains potassium hydroxide with a volume of 2.00 L and a concentration of 3.00 M. Take note of these values for future reference.

02

Write the balanced chemical equation

For this step, let's write the balanced chemical equation for the reaction between copper(II) nitrate (Cu(NO₃)₂) and potassium hydroxide (KOH): Cu(NO₃)₂(aq) + 2KOH(aq) → Cu(OH)₂(s) + 2KNO₃(aq) This equation tells us that one mole of copper(II) nitrate reacts with two moles of potassium hydroxide to form one mole of copper(II) hydroxide (precipitate) and two moles of potassium nitrate.

03

Calculate moles for each reactant

Calculate the moles of each reactant in Solution A and Solution B: For Cu(NO₃)₂: Moles = Volume × Concentration Moles = 2.00 L × 2.00 M = 4.00 moles For KOH: Moles = Volume × Concentration Moles = 2.00 L × 3.00 M = 6.00 moles

04

Determine the limiting reactant and the moles of the products formed

Based on the balanced chemical equation, one mole of Cu(NO₃)₂ reacts with two moles of KOH. Since there are 4 moles of Cu(NO₃)₂ and 6 moles of KOH, Cu(NO₃)₂ is the limiting reactant. Now, let's determine the number of moles of products formed: For Cu(OH)₂: Formed moles = moles of limiting reactant = 4.00 moles For KNO₃: Formed moles = 2 × moles of limiting reactant = 2 × 4.00 = 8.00 moles

05

Calculate concentrations of remaining ions in solution

The total volume of the mixed solutions is 4.00 L (2.00 L + 2.00 L). Let's calculate the final concentration of each ion in the solution: For K⁺ ions: Concentration = moles / total volume = 8.00 moles / 4.00 L = 2.00 M For NO₃⁻ ions: Concentration = (unreacted moles + formed moles) / total volume Unreacted moles of NO₃⁻ ions = initial moles in Solution A – moles reacted with KOH = 4.00 moles – 4.00 moles = 0 moles Concentration = (0 + 8.00 moles) / 4.00 L = 2.00 M OH⁻ ions completely react with Cu(NO₃)₂, so there are no OH⁻ ions left in the solution.

06

Calculate the mass of solid formed

To find the mass of copper(II) hydroxide precipitate, use the formula: Mass = moles × molar mass The molar mass of Cu(OH)₂ is 63.55 (Cu) + 2 × (16.00 + 1.01) = 97.57 g/mol Mass = 4.00 moles × 97.57 g/mol = 390.28 g In summary, after the precipitation reaction, the concentrations of K⁺ and NO₃⁻ ions in the solution are both 2.00 M, and the mass of Cu(OH)₂ solid formed is 390.28 g. For part a of the question, you can create a drawing comparing the relative volume of the mixed solution (which should be twice as large as Solution A or Solution B) and representing the copper(II) hydroxide solid formed, as well as the presence of K⁺ and NO₃⁻ ions in solution.

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