Concentration and dilution of aqueous solutions (QCE Chemistry Unit 2)

What this dot point is asking

QCAA wants you to express solution concentration in every standard unit (mol/L, g/L, %m/v, %m/m, ppm) and convert between them; use the dilution formula c_1 V_1 = c_2 V_2 confidently; and apply concentrations to stoichiometric calculations for reactions in aqueous solution. This dot point is heavily tested in EA Paper 1 and provides foundation for the titration calculations later in the year.

The answer

Concentration is the amount of solute per unit amount of solution (or solvent). Several units are in routine use; you need to convert fluently between them and apply each correctly in stoichiometry and dilution problems.

Standard concentration units

For dilute aqueous solutions where density is approximately 1.00 g/mL:

• 1 ppm = 1 mg/L = 1 mg/kg = 1 microgram/g.
• 1 ppb = 1 microgram/L = 1 microgram/kg.

If the solution density differs significantly from water, do the calculation rigorously (m/m and v/v are not interchangeable then).

Calculating molarity

where c is in mol/L, n in mol, V in L. Rearrangements:

• n = c V (moles of solute in a given volume of solution).
• V = n / c (volume needed to deliver a given amount).

Note: V must be litres. Convert mL by dividing by 1000.

Worked example: preparing a standard solution

A chemist needs 250.0 mL of 0.0500 mol/L silver nitrate solution.

Mass of AgNO_3 required: n = c V = 0.0500 x 0.2500 = 0.01250 mol. M(AgNO_3) = 169.87 g/mol. m = n x M = 0.01250 x 169.87 = 2.123 g.

Procedure: weigh 2.123 g of AgNO_3 accurately, dissolve in less than 250 mL of deionised water, transfer quantitatively to a 250.0 mL volumetric flask, make up to the mark, invert to mix.

Interconverting units

Mass concentration to molarity:

ppm to mol/L for dilute aqueous solutions (density 1.00 g/mL):

(The factor of 1000 converts mg/L to g/L; division by M converts g/L to mol/L.)

Example: a 0.500 mol/L glucose solution. M(glucose) = 180.16 g/mol.

Mass concentration: 0.500 x 180.16 = 90.08 g/L.

%m/v: 9.008 g per 100 mL, so 9.01 %m/v.

ppm (approximating density as 1.00 g/mL): 90,080 mg/L = 90,080 ppm. (At this concentration the density assumption is no longer accurate; do not push the approximation past about 1 percent solutions.)

Dilution

Diluting a solution increases the volume but does not change the amount of solute. Therefore concentration falls in inverse proportion to volume.

(Both c units must match and both V units must match; the units cancel.)

Two routine cases:

• Calculate how much stock to use: V_1 = c_2 V_2 / c_1.
• Calculate the resulting concentration after dilution: c_2 = c_1 V_1 / V_2.

Worked example: prepare 100.0 mL of 0.100 mol/L HCl from a 2.00 mol/L stock.

V_1 = (0.100 x 100.0) / 2.00 = 5.00 mL.

Pipette 5.00 mL of 2.00 mol/L HCl into a 100.0 mL volumetric flask. Add water to about three-quarters full, mix, then make up to the mark. (Adding stock to water reduces splash/heat risk for strong acids; for HCl this is not a concern but is good practice for H_2SO_4 dilutions.)

Solution stoichiometry

Solution stoichiometry uses the same mole map as gas stoichiometry, with V x c replacing V_m or PV/(RT) at the solution end.

Steps:

1. Convert known solution volume and concentration to moles: n = c V.
2. Apply the mole ratio from the balanced equation.
3. Convert moles to the required final unit (mass, volume of solution, gas volume).

Worked example: 25.00 mL of 0.1000 mol/L NaOH is required to neutralise an unknown HCl solution. 18.50 mL of HCl is used. Find c(HCl).

n(NaOH) = c V = 0.1000 x 0.02500 = 2.500 x 10^-3 mol.

NaOH + HCl -> NaCl + H_2O, mole ratio 1 : 1, so n(HCl) = 2.500 x 10^-3 mol.

c(HCl) = n / V = (2.500 x 10^-3) / 0.01850 = 0.1351 mol/L.

This is the standard titration calculation; the full titration dot point is treated separately in Unit 3.

Serial dilution

A serial dilution is a sequence of dilutions, each one applied to the previous result. The total dilution factor is the product of the individual dilution factors.

Example: 1 mL of stock diluted to 10 mL (factor 10), 1 mL of that diluted to 100 mL (factor 100). Total dilution factor = 10 x 100 = 1000. Final concentration = c_stock / 1000.

Serial dilution is used for low-concentration standards (calibration curves) and biological work.

Examples in context

Example 1. Standardised pesticide dosing at Mackay sugar farms. Cane growers in the Mackay-Whitsunday region prepare diuron herbicide at $1.50 \, \text{g L}^{-1}$ for boom-spray application. To make $200 \, \text{L}$ from a stock $50 \, \text{g L}^{-1}$ concentrate, operators apply $c_1 V_1 = c_2 V_2$: $50 \times V_1 = 1.50 \times 200$, so $V_1 = 6.00 \, \text{L}$ of stock topped to $200 \, \text{L}$ with water. Getting the dilution right is critical because Great Barrier Reef Marine Park Authority runoff limits cap diuron at $0.43 \, \mu \text{g L}^{-1}$ in waterways, and over-application is a frequent IA2 case-study scenario for Queensland chemistry students.

Example 2. Hospital saline at the Townsville University Hospital pharmacy. Pharmacists prepare $0.9\%$ m/v saline by dissolving $9.00 \, \text{g}$ NaCl per $1.00 \, \text{L}$ sterile water, giving a molarity of $n/V = (9.00 / 58.44) / 1.00 = 0.154 \, \text{mol L}^{-1}$. To prepare $500 \, \text{mL}$ of this from a $5.00 \, \text{mol L}^{-1}$ concentrated stock, the dilution formula gives $V_1 = (0.154 \times 0.500) / 5.00 = 0.0154 \, \text{L} = 15.4 \, \text{mL}$. The remainder is sterile water of injection. Accurate dilutions matter clinically because osmolarity must match blood serum to avoid haemolysis.

Try this

Q1. Calculate the concentration in $\text{mol L}^{-1}$ when $2.50 \, \text{g}$ of $\text{NaOH}$ ($M = 40.00$) is dissolved in $250 \, \text{mL}$ of water. [3 marks]

• Cue. $n = 2.50/40.00 = 0.0625 \, \text{mol}$; $c = 0.0625 / 0.250 = 0.250 \, \text{mol L}^{-1}$.

Q2. $25.0 \, \text{mL}$ of $0.500 \, \text{mol L}^{-1}$ HCl is diluted to $250 \, \text{mL}$. Calculate the final concentration. [3 marks]

• Cue. $c_1 V_1 = c_2 V_2$: $c_2 = (0.500 \times 25.0)/250 = 0.0500 \, \text{mol L}^{-1}$. Tenfold dilution.

Q3. A student must prepare $250 \, \text{mL}$ of $0.100 \, \text{mol L}^{-1}$ $\text{KMnO}_4$ from solid. (a) Calculate the mass needed. (b) Describe the volumetric procedure. (c) Identify one source of random error. [3+3+1 marks]

• Cue. (a) $0.100 \times 0.250 \times 158.04 = 3.95 \, \text{g}$. (b) Dissolve in beaker; transfer to vol flask with washings; make to mark. (c) Meniscus reading. Communication ISMG.