6.4 Aqueous Solutions

An aqueous solution is one in which the solvent is water.

Salt and water is an example of aqueous solutions where the solute is a solid.

Alcohol and water is an example of aqueous solutions where the solute is a liquid.

Ammonia and water is an example of aqueous solutions where the solute is a gas

Concentration: relative amounts of solute and solvent.

Molar concentration (Molarity): is the number of moles of solute per liter (dm3) of

solution. (the relative amounts of solute and solution)

0.5 M of solution means 0.5 moles of solute in 1 dm3 solution

1.0 M of solution means 1.0 moles of solute in 1 dm3 solution

5.0 M of solution means 5.0 moles of solute in 1 dm3 solution

Concentration of a given solution does not change if solution is split into fractions.

Relationships between n, V, C and m, M, V, C:

Preparing solutions with given concentrations

Dilution of solution: If a solution is diluted by adding water 

Number of moles in concentrated solution = number of moles of dilute solution. CV = C’V’

 Upon dilution, volume of water needed for dilution = V dilute solution – V concentrated

solution

1) Pour the 100 cm3 of 4.0 M salt solution into a clean volumetric 500 cm3 flask. Add

enough water to fill the flask up to the etched mark. What is the concentration of the salt

solution in the new flask?

2) 200 cm3of 0.40 M NaCl solution was poured into a 500 cm3 beaker. Water was added till

the etched mark.

a) What is the new concentration?

b) What is the volume of water needed for dilution?

3) You are given 100 cm3 of 0.50 M HCl, how much water must you add to reduce the

concentration to 0.10M.

4) To what volume must 50.0 cm3 of 3.50 M H2SO4 be diluted in order to make a 2.00 M

H2SO4?

To find [solutes] after mixing two or more solutions:

Step 1: Calculate the number of moles of each solution

Step 2: Calculate the total volume

Step 3: Find new concentration.

Mixing equal volumes of different solutions halves the concentration of all species in

the solution.

Saturated solution: is a solution that has dissolved in it the maximum amount of solute

it can hold at a given temperature and is in contact with the undissolved solute. In simple

words, a saturated solution is a solution in which no more solute can dissolve.

Solubility: it is the concentration of a saturated solution at a certain temperature

Soluble solubility > 0.1 M

Slightly soluble solubility < 0.1 M

Very slightly soluble solubility < 10-3 M

Insoluble or negligible solubility solubility is low

SELF STUDY: the electric model of atoms, direction of electric current, meaning of

fundamental property, effect of distance on electric force.

Molar Volume in the Solid and Liquid Phase

• The solid and liquid phases for a particular substance typically have similar molar volume

  • This is because the distance between the particles of the substance in the solid phase is approximately the same as the distance between the particles in the liquid phase

  • Molar volume refers to the volume of one mole of a substance under standard conditions

Table of Molar volumes for metals in solid and liquid states

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:

Exam Tip

• 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

Exam Tip

• 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

Molarity

• A solution may be described qualitatively or quantitatively

  • Qualitatively, the terms dilute and concentrated are used to describe the amount of solute in a given quantity of the solvent

    • A solution with a relatively small concentration of solute is said to be dilute

    • A solution with a large concentration of solute is said to be concentrated

  • Quantitatively, a solution is described in terms of its concentration

Molarity

• Molarity (M) expresses the concentration of a solution as the number of moles of solute in a litre of solution:

M = n / V

• A 1.00 molar solution (1.00 M) contains 1.00 mol of solute in every 1.0 L of solution

• A solution can be prepared to a specified molarity by weighing out the calculated mass of solute and dissolving it in enough solvent to form the desired volume of solution

• For example, to prepare 250.0 mL of a 1.00 M solution of CuSO4:

  • Determine the number of moles of the solute (CuSO4) required

    • In this case, this is given as n = Molarity × Volume

      • n = 1.00 × 0.250  = 0.250 mol

  • Determine and weigh the corresponding mass of CuSO4 required

    • Using the the expression: Mass = n × Mr

      • The molar mass (Mr) of CuSO4 is calculated as:

        • Mr = (Mr)cu + (Mr)S + 4(Mr)O

        • Mr = 63.6 + 32.0 + 4 (16) = 159.6 g/mol

      • Mass = 0.250 mol × 159.6 g/mol = 39.9 g

    • Transfer 39.9 g of CuSO4 into a 250 mL flask and add some quantity of water to dissolve the solute

    • Add more water until the solution reaches the calibrated mark of the flask

Preparing a Solution

How to prepare a 250mL of 1.00 M solution of CuSO4

Dilution

• Alternatively, you can start with a more concentrated solution, called the stock solution, and dilute it with water to give a solution of the desired molarity

  • The calculations are straightforward if you keep a simple point in mind: Adding solvent cannot change the number of moles of solute

    • That is:

nsolute (stock solution) =  nsolute (dilute solution)

• In both solutions, n can be found by multiplying the molarity, M, by the volume in litres, V

• Hence, a dilution expression can be obtained:

Ms × Vs = Md × Vd

• • Where the subscripts s and d stand for stock and dilute solutions, respectively

Worked example

Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution. (Mr glucose = 180.2 g/mol)

Answer:

Analyse: We are provided with the mass of glucose and the volume of solution and asked to determine the molarity of the solution

Plan: To determine the molarity of the solution the following steps are required:

Step 1: Convert the mass of glucose to moles and the volume of the solution to litres

• n = Mass/Molar Mass        

• n = 5.00/180.2                    

• n = 0.0277 moles

• Volume = 100/1000

• Volume = 0.1L

Step 2: Substitute the values of the moles of glucose and the volume of the solution

• Molarity (M) = number of moles/volume of solution

• Molarity (M) = 0.0277/0.1

• Molarity (M) = 0.277 M

Worked example

1. How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4?

2. How many millilitres of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt?

Analyse:

• In part a, we are given the volume of the solution and the concentration (Molarity)

  • We are then asked to determine the mass of the solute(Na2SO4) required

• In part b, we are given the number of moles of solute and the molarity of the solution

  • We are then asked to determine the volume of solution in millilitres

Plan:

For part a:

• Use the molarity expression to determine the number of moles of the solute (n)

• Convert this number of moles to mass

For part b:

• Use the molarity expression to determine the volume of solution in litres

• Convert to millilitres using the appropriate conversion factor

Solution

• Part a:

  • Step 1: Convert volume in mL to L:

    • Volume of solution = 15/1000 = 0.015L

  • Step 2: Rearrange the concentration expression in terms of the number of Na2SO4 (n):

    • n = M × V

    • n = 0.50 × 0.015

    • n = 0.0075 moles

  • Step 3: Using the mole-molar mass expression, convert the moles of solute to mass:

    • MNa2SO4 = n × Mr

    • MNa2SO4 = 0.0075 × 142.04

    • MNa2SO4 = 1.1g

• Part b:

  • Step 1: Rearrange the concentration expression in terms of the volume of solution:

    • V = n/M

    • V = 0.038/0.50

    • V = 0.076 L

  • Step 2: Convert volume in L to mL:

    • V = 0.076 × 1000

    • V = 76 mL

Worked example

How many millilitres of 5.0 M Na2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? How much water needs to be added to this concentrated solution?

Analyse:

• We are given the volume and concentration of the diluted solution and asked to determine the volume of the concentrated solution required to prepare

• We are then asked the volume of water required to make the dilute solution

Plan:

• Using the dilution formula, we can calculate the volume of the concentrated solution required

• Then subtract that volume from the volume of the dilute solution to obtain the volume of water required

Answer:

• Step 1: Using M1V1 = M2V2

  • Rearrange the equation to V1 = (M2V2)/ M1

    • V1 = ?

    • M1 = 5.0 M

    • V2 = 250 mL

    • M2 = 0.10 M

  • So, V1 = (0.10 × 250)/5.0

  • V1 = 5.0 mL

• Step 2: The volume of water added will be:

  • VH2O = 250 - 5.0

  • VH2O = 245 mL

Particulate Representations of Solutions

• A solution is an homogeneous mixture of two substances

• The components of a solution are a solute and a solvent

  • The solvent is the component that is in the highest amount

    • Liquids usually act as solvents in solutions

  • The solute is the component that is in the least amount

• E.g.  A mixture of salt water is the best example for a solution. Salt is the solute and water acts as a solvent

• Any solution in which water acts as a solvent is called an aqueous solution

• Concentration makes reference to the amount of particles of solvent into an specific volume of solute

  • The greater the amount of solute particles in a fixed volume, the greater the concentration

• A concentrated solution has a large amount of solute particles

• A diluted solution has a relatively small amount of solute

Answer:

The correct answer is C because:

• The particles of solute and solvent are evenly spread in the container

• It has a largest amount of solute particles (salt particles) compared to B which is also a solution

• A and D are not correct. The particles are forming individual regions of solute and solvent. Therefore, they are heterogenous mixtures which are not solutions

1) Salt and water. (a solution of solid in water) 

2) Alcohol and water. (a solution of liquid in water)

3) Carbon dioxide and water. (a solution of gas in water)