Calculations About Solutions (College Board AP® Chemistry)

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Oluwapelumi Kolawole

Written by: Oluwapelumi Kolawole

Reviewed by: Stewart Hird

Calculations About Solutions

  • Usually in calculations involving solutions, we are required to determine:

    • Number of moles or mass of solute

    • Number of moles of ions that make up the solute

    • Molarity of solution

    • Volume of solution

  • The calculation would involve dilution and require the use of the dilution formula

    • Dilution formula: Ms × Vs = Md × Vd

    • This is different from calculations involving mixing two solutions containing the same solute or with a common ion and being asked to determine the concentration of the mixed solution or the common ion

Worked Example

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:

    • begin mathsize 16px style straight n subscript Pb left parenthesis NO subscript 3 right parenthesis subscript 2 end subscript space equals space straight n subscript Pb to the power of 2 plus end exponent end subscript space equals space 0.50 space mol end style

    • straight n subscript NO subscript 3 to the power of minus end subscript space equals space 2 space cross times space straight n subscript Pb to the power of 2 plus end exponent end subscript space equals space 2 space cross times space 0.50 space equals 1.0 space 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 = 0.125 L or 125 mL

Examiner Tips and Tricks

  • This overall question contains more than one question but is not split into parts a, b, etc

    • When you encounter questions like this, you do not have to answer the questions in the order they are asked

  • For this question, you cannot answer the first actual question before the second one because the number of moles of the solute, Pb(NO3)2, is required to calculate the volume of the solution

Worked Example

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:

      • nKOH = M × V (Remember volume must be in L)

      • nKOH = 0.1 × 0.0550

      • nKOH = 0.0055 mol

    • For solution B:

      • nKOH = M × V (Remember volume must be in L)

      • nKOH = 0.15 × 0.0750

      • nKOH = 0.01125 mol

  • Step 2: Add up the number of moles of KOH from each solution:

    • (nKOH)T = (nKOH)A + (nKOH)B

    • (nKOH)T = 0.0055 + 0.01125

    • (nKOH)T = 0.01675 mol

  • Step 3: Determine the total volume of the mixture, in L:

    • VT = 75.0 + 55.0

    • VT = 130.0/1000

      • Remember: volume must be in L

    • VT = 0.130 L

  • Step 4: Determine the molarity of the mixture by dividing the total number of moles by the total volume:

    • MT = nT/VT

    • MT =0.01675/0.130

    • MT = 0.129 M

Worked Example

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 (Vd):

    • Vd = Ms × Vs / Md

    • Vd = 2.4 × 500.0 / 1.0

    • Vd = 1200 mL

  • Step 2: Subtract the volume of the stock solution (Vs) from the volume of the dilute solution (Vd) to obtain the volume of water:

    • VH2O = Vd - Vs

    • VH2O = 1200 - 500

    • VH2O = 700 mLs

Examiner Tips and Tricks

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

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

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Oluwapelumi Kolawole

Author: Oluwapelumi Kolawole

Expertise: Chemistry Content Creator

Oluwapelumi is a Pharmacist with over 15000+ hours of AP , IB, IGCSE, GCSE and A-Level chemistry tutoring experience. His love for chemistry education has seen him work with various Edtech platforms and schools across the world. He’s able to bring his communication skills as a healthcare professional in breaking down seemingly complex chemistry concepts into easily understood concepts for students.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.