Calculations About Solutions (College Board AP Chemistry)

What volume of 4.0 M lead(II) nitrate solution, Pb(NO₃)₂ (aq), contains 0.50 mol of Pb²⁺? What is the number of moles of NO₃^- ions present in this solution?

Analyse:

• We are provided with the concentration of the solution (4.0 M) and the number of moles of Pb²⁺ ions (0.50 mol)
• We are required to determine the volume of the solution and the number of moles of the nitrate ions

Plan:

• First, we determine the number of moles of the Pb(NO₃)₂ solute from the number of moles of Pb²⁺ ions using the mole ratio from the dissociation ionic equation
• Use the same mole ratio to determine the number of moles of nitrate, NO₃^-, ions
• Then use the calculated number of moles of the solute and the concentration to determine the volume of the solution

Answer:

• Step 1: Write the dissociation equation for the solute:
  • Pb(NO₃)₂ (aq) → Pb²⁺ (aq) + 2 NO₃^- (aq)
• Step 2: From the equation, the mole ratio of Pb(NO₃)₂ : Pb²⁺ is 1:1 and Pb²⁺: NO₃^- is 1:2, then:
  • 
  • 
• Step 3: Determine the volume of the solution using the molarity expression
  • n = M × V
  • V = n/M
  • V = 0.50/4.0
  • V = 0.125 L or 125 mL

What is the molar concentration of the KOH solution obtained from mixing two solutions of KOH, A and B?

• Solution A: 55.0 mL of 0.10 M KOH
• Solution B: 75.0 mL of 0.15 M KOH

Analyse: We are provided with the concentration and volume of two aqueous KOH solutions and asked to determine the concentration of the solution on mixing

Plan:

• Determine the number of moles of the solute in each solution using the molarity equation
• Determine the total number of moles of the solute by adding the number of moles from each solution
• Divide this total number of solute moles by the total volume of the solution

Answer:

• Step 1: Determine the number of moles of KOH in solutions A and B:
  • For solution A:
    • n_KOH = M × V (Remember volume must be in L)
    • n_KOH = 0.1 × 0.0550
    • n_KOH = 0.0055 mol
  • For solution B:
    • n_KOH = M × V (Remember volume must be in L)
    • n_KOH = 0.15 × 0.0750
    • n_KOH = 0.01125 mol
• Step 2: Add up the number of moles of KOH from each solution:
  • (n_KOH)_T = (n_KOH)_A + (n_KOH)_B
  • (n_KOH)_T = 0.0055 + 0.01125
  • (n_KOH)_T = 0.01675 mol
• Step 3: Determine the total volume of the mixture, in L:
  • V_T = 75.0 + 55.0
  • V_T = 130.0/1000
    • Remember: volume must be in L
  • V_T = 0.130 L
• Step 4: Determine the molarity of the mixture by dividing the total number of moles by the total volume:
  • M_T = n_T/V_T
  • M_T =0.01675/0.130
  • M_T = 0.129 M

How much water would be required to dilute 500.0 mL of 2.4 M CaCl₂ solution to make a 1.0 M solution?

Analyse: We are asked to determine the quantity of water required to be added to a stock solution (500 mL, 2.4 M) to give a 1.0 M dilute solution

Plan:

• Using the dilution formula, we determine the volume of the dilute solution
• Subtract the volume of the stock solution from the final volume to obtain the volume of water added

Answer:

• Step 1: Rearrange the dilution formula to determine the volume of the dilute solution (V_d):
  • V_d = M_s × V_s / M_d
  • V_d = 2.4 × 500.0 / 1.0
  • V_d = 1200 mL
• Step 2: Subtract the volume of the stock solution (V_s) from the volume of the dilute solution (V_d) to obtain the volume of water:
  • V_H2O = V_d - V_s
  • V_H2O = 1200 - 500
  • V_H2O = 700 mLs