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Solutions: concentration, preparation and ionic strength

Almost all laboratory chemistry happens in solution. This topic explains the ways of expressing a concentration, how to go from one to another (and when the density is needed), how to prepare and dilute a solution, and what ionic strength is, the measure of how far a solution of electrolytes departs from ideal behavior.

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Ways of expressing concentration

A solution has a solute (what dissolves) and a solvent (almost always water). The concentration says how much solute there is, but it can be measured in many ways, each useful for something:

Measure	Definition	Used for
Molarity $M$	mol of solute / L of solution	preparing and titrating (volumes)
Molality $b$	mol of solute / kg of solvent	colligative properties (does not change with T)
%m/m	g of solute / 100 g of solution	commercial concentrated reagents
%m/v	g of solute / 100 mL of solution	serums and clinical solutions
%v/v	mL of solute / 100 mL of solution	mixtures of liquids (alcohol)
ppm	mg of solute / kg of solution	traces, waters, environment
Mole fraction $x$	mol of solute / total mol	gases, vapor pressure

The measures «per volume of solution» (M, %m/v) and the ones «per mass» (b, %m/m, ppm, x) are only connected through the density of the solution, $ρ = m_{so l} / V_{so l}$ . Without it you cannot go, for example, from molarity to molality. In a very dilute aqueous solution one can take $ρ \approx 1.00$ g/mL (that of water), and then 1 ppm ≈ 1 mg/L; but that only holds when the solute is a very small fraction of the mass.

M = \frac{n _{s t o}}{V _{so l}} b = \frac{n _{s t o}}{m _{s t e}} m_{so l} = ρ V_{so l} = m_{s t o} + m_{s t e} x_{s t o} = \frac{n _{s t o}}{n _{s t o} + n _{s t e}}

If the problem only gives concentrations (98% m/m and a density), no amount is fixed: you choose a basis of calculation (1 L or 100 g of solution), compute the masses and moles of that basis, and the concentrations obtained hold for any amount.

Preparing and diluting

To prepare a volume $V$ of molarity $M$ from a solid you weigh $m = M \cdot V \cdot M$ (with $M$ the molar mass), dissolve it in part of the water inside a volumetric flask and fill with water up to the mark. The flask is filled at the end because the volume of the mixture is not the sum of the volumes.

Preparation in a volumetric flask: 500 mL of 0.1 M NaOH (example 2).

When diluting solvent is added and the moles of solute do not change, so $M_{1} V_{1} = M_{2} V_{2}$ : to get $V_{2}$ of concentration $M_{2}$ you take $V_{1} = M_{2} V_{2} / M_{1}$ of the concentrated solution and bring it to $V_{2}$ . With concentrated acids the acid is added to the water, never the other way around: the mixing releases a lot of heat.

Ionic strength and activity

In a solution of electrolytes each ion is surrounded by an «atmosphere» of ions of opposite charge that screens it, and it behaves as if there were less of it. That effective concentration is the activity, $a = γ c$ , with an activity coefficient $γ \leq 1$ . The effect depends on the ionic strength, in which each ion counts with the square of its charge:

I = \frac{1}{2} i \sum c_{i} z_{i}^{2} lo g γ = - A z^{2} I lo g γ = - A z^{2} \frac{I}{1 + I} lo g γ = - A z^{2} (\frac{I}{1 + I} - 0.3 I)

They are, in order, the Debye–Hückel limiting law (valid up to $I \approx 0.005$ M), Güntelberg's (up to 0.1 M) and Davies's (up to about 0.5 M), with $A = 0.509$ for water at 25 °C. An ion of charge 2 departs four times more (in $lo g γ$ ) than one of charge 1, and one of charge 3, nine times.

Activity coefficient versus ionic strength, with each equation over its range of validity.

Common mistakes

Dividing by the volume of solvent instead of solution (molarity) or by the mass of solution instead of solvent (molality).
Taking $ρ = 1$ g/mL for a concentrated solution: 98% sulfuric acid has 1.84 g/mL.
Adding solute and solvent volumes to get the solution's volume.
Forgetting the subscript when dissociating: Na₂SO₄ gives 2 Na⁺ per formula.
Computing the ionic strength with the charge not squared, or without the ½.

Worked examples

Example 1Molarity from the mass

5.85 g of NaCl are dissolved and made up to 250 mL of solution. Find the molarity and the % m/v. Can the molality be known?

Solution

$M (NaCl) = 22.990 + 35.45 = 58.44$ g/mol, so $n = 5.85/58.44 = 0.1001$ mol.
$M = 0.1001/0.250 = 0.4004$ M and %m/v $= 5.85/250 \cdot 100 = 2.34%$ .
For the molality you need the mass of water, and for that the density of the solution, which is not given. With 2.34% solute, assuming $ρ = 1.00$ g/mL is no longer safe, so the molality is left uncalculated.

M = 0.4004

M, 2.34% m/v; the molality needs the density.

Solve it in Simuveliq

In Chemistry → Prepare a solution, in Examples, «5.85 g of NaCl in 250 mL of solution». The Concentrations card gives M = 0.4004, %m/v = 2.34% and 23 400 mg/L, and the Step by step, each formula.
The molality does not appear: the tool does not use ρ ≈ 1.00 g/mL because the solute exceeds 0.1% (with «Dilute-solution approximation» = «Use it only if needed and valid»). Type a density and it appears.

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Example 2How much to weigh for a solution

How many grams of NaOH must be weighed to prepare 500 mL of a 0.1 M solution?

Solution

Moles needed: $n = M \cdot V = 0.1 \cdot 0.500 = 0.05$ mol.
$M (NaOH) = 22.990 + 15.999 + 1.008 = 39.997$ g/mol, so $m = 0.05 \cdot 39.997 = 1.9998$ g ≈ 2.000 g.
They are weighed, dissolved in about 300 mL of water in a 500 mL flask (the solution warms up), left to cool and filled up to the mark.

Weigh 1.9998 g (≈ 2.000 g) of NaOH.

Solve it in Simuveliq

«How much NaOH to weigh for 500 mL at 0.1 M?»: with What do you want to know? = mass of solute, the tool computes n = 0.05 mol and msto = 2 g (to 4 figures; with the molar mass 39.9968 g/mol shown below, 1.9998 g).

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Example 3Concentrated sulfuric acid: M, b and dilution

Commercial H₂SO₄ is 98% m/m with a density of 1.84 g/mL. Find its molarity, molality and mole fraction, and the volume needed to prepare 1 L of 1 M H₂SO₄.

Solution

Basis of 1 L: $m_{so l} = 1.84 \cdot 1000 = 1840$ g, of which $0.98 \cdot 1840 = 1803.2$ g are acid and $36.8$ g water.
$n = 1803.2/98.072 = 18.39$ mol, so $M = 18.39$ M and $b = 18.39/0.0368 = 499.6$ mol/kg. Water: $36.8/18.015 = 2.043$ mol → $x = 18.39/ (18.39 + 2.043) = 0.900$ .
Dilution to 1 M: $V_{1} = M_{2} V_{2} / M_{1} = 1 \cdot 1000/18.39 = 54.4$ mL of concentrated acid, added slowly to water, then made up to 1 L.

M = 18.39

b = 499.6

mol/kg,

x = 0.900

; for 1 L of 1 M, 54.4 mL are taken.

Solve it in Simuveliq

«98% m/m H₂SO₄, ρ = 1.84 g/mL: M and molality»: the tool says it takes 1 L as the basis, and gives M = 18.39, b = 499.6 mol/kg, x = 0.9 and the step by step of the basis of calculation.

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Example 4Every concentration of a solution

20 g of KCl are dissolved in 180 g of water; the solution has a density of 1.08 g/mL. Find M, b, % m/m and mole fraction.

Solution

$m_{so l} = 200$ g → %m/m = 10% and $V_{so l} = 200/1.08 = 185.19$ mL.
$n = 20/74.548 = 0.2683$ mol → $M = 0.2683/0.18519 = 1.449$ M and $b = 0.2683/0.180 = 1.490$ mol/kg.
Water: $180/18.015 = 9.992$ mol → $x = 0.2683/ (0.2683 + 9.992) = 0.02615$ .

M = 1.449

b = 1.490

mol/kg, 10% m/m,

x = 0.02615

Solve it in Simuveliq

«20 g of KCl in 180 g of water, ρ = 1.08 g/mL» gives every concentration, the solution volume (185.2 mL) and each step.

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Example 5Ionic strength of a mixture

A solution has 0.1 M NaCl, 0.05 M CaCl₂ and 0.02 M Na₂SO₄. Find the ionic strength and the activity coefficients.

Solution

Ions: Na⁺ $0.1 + 2 \cdot 0.02 = 0.14$ M; Cl⁻ $0.1 + 2 \cdot 0.05 = 0.2$ M; Ca²⁺ 0.05 M; SO₄²⁻ 0.02 M. The charges balance: $0.14 + 2 (0.05) = 0.2 + 2 (0.02) = 0.24$ .
$I = \frac{1}{2} (0.14 \cdot 1 + 0.2 \cdot 1 + 0.05 \cdot 4 + 0.02 \cdot 4) = \frac{1}{2} \cdot 0.62 = 0.31$ M.
With 0.31 M Davies is used ( $I = 0.5568$ ): for $z = 1$ , $lo g γ = - 0.509 (0.5568/1.5568 - 0.093) = - 0.1347$ → $γ = 0.733$ ; for $z = 2$ , four times the logarithm → $γ = 0.289$ .

I = 0.31

γ = 0.733

(Na⁺, Cl⁻) and 0.289 (Ca²⁺, SO₄²⁻) with Davies.

Solve it in Simuveliq

In Chemistry → Ionic strength, example «0.1 M NaCl + 0.05 M CaCl₂ + 0.02 M Na₂SO₄». It gives I = 0.31 M, √I = 0.5568, «charges balanced ✓», the ion table with c·z² and the coefficients with the three equations (Davies: 0.7333 and 0.2892).

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Example 6An ion of charge 3

1 g of Al₂(SO₄)₃ is dissolved in 250 mL. Find the ionic strength and the activity coefficient of Al³⁺.

Solution

$M = 342.131$ g/mol → $c = (1/342.131) /0.25 = 0.01169$ M. Ions: Al³⁺ $2 c = 0.02338$ M and SO₄²⁻ $3 c = 0.03507$ M.
$I = \frac{1}{2} (0.02338 \cdot 9 + 0.03507 \cdot 4) = 0.1754$ M: fifteen times the solute concentration, because the charges are high.
Davies with $z = 3$ : $lo g γ = - 0.509 \cdot 9 (0.4188/1.4188 - 0.0526) = - 1.111$ → $γ = 0.0774$ . The aluminum acts as if there were less than a tenth of it.

I = 0.1754

γ (Al^{3 +}) = 0.0774

(Davies).

Solve it in Simuveliq

Example «1 g Al₂(SO₄)₃ in 250 mL»: step 1 goes from grams to molarity, the ion table gives c·z² = 0.2104 and 0.1403, and I = 0.1754 M; in the coefficient table, Davies gives 0.07742 for Al³⁺.

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Solutions: concentration, preparation and ionic strength

Ways of expressing concentration

Preparing and diluting

Ionic strength and activity

Common mistakes

Worked examples

Example 1Molarity from the mass

Solution

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Example 2How much to weigh for a solution

Solution

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Example 3Concentrated sulfuric acid: M, b and dilution

Solution

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Example 4Every concentration of a solution

Solution

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Example 5Ionic strength of a mixture

Solution

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Example 6An ion of charge 3

Solution

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