Calculations About Solutions (College Board AP® Chemistry): Revision Note

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?

Analyze:

• The concentration of the solution (4.0 M) and the number of moles of Pb²⁺ ions (0.50 mol) are given

• Find the volume of the solution and the number of moles of nitrate ions

Plan:

• First, find the number of moles of Pb(NO₃)₂ from the number of moles of Pb²⁺ using the mole ratio in the dissociation 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:

  • ${\mathrm{n}}_{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2}={\mathrm{n}}_{{\mathrm{Pb}}^{2+}}=0.50\mathrm{mol}$

  • ${\mathrm{n}}_{{{\mathrm{NO}}_3}^{-}}=2\times {\mathrm{n}}_{{\mathrm{Pb}}^{2+}}=2\times 0.50=1.0\mathrm{mol}$

• Step 3: Determine the volume of the solution using the molarity expression

  • n = M × V

  • V = n/M

  • V = 0.50/4.0

  • V = 0.13 L (130 mL) (to 2 significant figures)

• Read the formula before calculating ion amounts: 1 mol of Pb(NO₃)₂ gives 1 mol of Pb²⁺ and 2 mol of NO₃^-, so n(NO₃^-) = 2 × n(Pb(NO₃)₂)

• Molarity is not moles: n = M × V, with V in liters (multiply, never divide, M by V)

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

Analyze: The concentration and volume of two aqueous KOH solutions are given; find the concentration of the solution formed 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 = 0.10 × 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 = 0.13 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?

Analyze: Find the volume of water that must be added to the stock solution (500.0 mL, 2.4 M) to give a 1.0 M solution

Plan:

• Use the dilution formula to find the total 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 mL (7.0 × 10² mL to 2 significant figures)

• When using the dilution formula, ensure the volume of the dilute and stock solutions are in the same units

• In the dilution formula, V_d is the TOTAL volume of the diluted solution, not the volume of water added. Find the water added afterwards: V_d − V_s (as in the worked example above)

• When using the molarity expression, always change your volume to liters, L