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Species present and their relative concentrations

What the question asks

You are given a solution and must list every species present, in order of decreasing concentration.
This tests whether you can see what actually happens when a substance dissolves, and it appears on the paper most years.
Water is never listed — it is the solvent, and it is present in vast excess.

The method

Write what the substance gives on dissolving.
Decide which ion, if any, reacts with water — it is always the one derived from the weak partner.
Write that equilibrium and identify the species it produces.
Rank them by concentration.

Which salts react with water

Salt	From	Ion that reacts	Solution
$\text{NaCl}$	strong acid + strong base	neither	neutral
$\text{CH}_3\text{COONa}$	weak acid + strong base	$\text{CH}_3\text{COO}^-$	basic
$\text{NH}_4\text{Cl}$	strong acid + weak base	$\text{NH}_4^+$	acidic

$\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$
$\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$
Ions from strong acids and bases — $\text{Cl}^-$ , $\text{NO}_3^-$ , $\text{Na}^+$ , $\text{K}^+$ — are spectators and do not react.

Ranking the species

Three principles get the order right every time:

The spectator ion is at its full concentration, because none of it is used up.
The ion that reacts is slightly below that, because a little has been converted.
The two products of that reaction are equal to each other, and both are small.
The ion that is not produced — $\text{H}_3\text{O}^+$ in a basic solution, $\text{OH}^-$ in an acidic one — is the smallest, tied to the others by $K_w$ .

Worked orderings

$0.10$ mol L⁻¹ $\text{CH}_3\text{COONa}$ — basic:

$[\text{Na}^+] > [\text{CH}_3\text{COO}^-] > [\text{CH}_3\text{COOH}] = [\text{OH}^-] > [\text{H}_3\text{O}^+]$

$0.10$ mol L⁻¹ $\text{CH}_3\text{COOH}$ — acidic, no spectator:

$[\text{CH}_3\text{COOH}] > [\text{H}_3\text{O}^+] = [\text{CH}_3\text{COO}^-] > [\text{OH}^-]$

$0.10$ mol L⁻¹ $\text{NH}_4\text{Cl}$ — acidic:

$[\text{Cl}^-] > [\text{NH}_4^+] > [\text{H}_3\text{O}^+] = [\text{NH}_3] > [\text{OH}^-]$

Note the equals signs. They are part of the answer, and questions often print an "=" between two of the boxes to tell you where they belong.

Linking concentration to conductivity

Conductivity depends on the total concentration of ions, whatever those ions are.
A salt solution is fully ionic, so it conducts well even if it is not acidic.
A weak acid is barely dissociated, so despite being acidic it conducts poorly.
So a solution's pH and its conductivity answer different questions, and a good answer uses each for what it shows.

Tips and Tricks

Never list water. It is the solvent and is present in enormous excess, so it does not belong in a ranking of solute species.

Put the equals sign in where two species are produced together. In a sodium ethanoate solution $[\text{CH}_3\text{COOH}] = [\text{OH}^-]$ exactly, because the reaction makes one of each.

The spectator ion is the largest, not the reacting one. Students often put the reacting ion first out of habit; it is very slightly smaller because some of it has been consumed.

Do not forget the minority ion. Even a strongly basic solution contains $\text{H}_3\text{O}^+$ — it is the smallest species, but it is there and it should be last on your list.

For a conductivity question, talk about the total ion concentration. Do not equate good conductivity with acidity.

Worked ExampleIdentifying three solutions from two measurements

Three colourless $0.110$ mol L⁻¹ solutions — $\text{CH}_3\text{NH}_2$ , $\text{CH}_3\text{COOH}$ and $\text{NH}_4\text{Cl}$ — have lost their labels and are relabelled A, B and C. Their conductivities and their colours with bromothymol blue ( $\text{p}K_a = 7.2$ , yellow in acid, blue in base) are:

Solution	Conductivity	Colour
A	poor	yellow
B	poor	blue
C	good	yellow

Identify each solution, justifying your answer in terms of the degree of dissociation and the relative concentration of ions, and give relevant equations. No calculations are necessary.

Step 1 — Classify each substance

$\text{CH}_3\text{NH}_2$ — methylamine, a weak base. Only partly reacts with water, so few ions.
$\text{CH}_3\text{COOH}$ — ethanoic acid, a weak acid. Only partly dissociates, so few ions.
$\text{NH}_4\text{Cl}$ — an ionic salt. It is fully dissociated into $\text{NH}_4^+$ and $\text{Cl}^-$ the moment it dissolves, so many ions.

Step 2 — Use the conductivity to find the salt

Conductivity depends on the total concentration of ions.

Only one solution conducts well, and only one substance is a fully ionised salt.

Its yellow colour with bromothymol blue confirms this is consistent — the solution should be acidic, because the ammonium ion is a weak acid:

$\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$

The chloride ion is the conjugate base of a strong acid and is a spectator.

Step 3 — Use the indicator colour to separate the two weak electrolytes

A and B both conduct poorly, so both are the weakly dissociated substances — the weak acid and the weak base. The indicator separates them.

A is yellow → acidic → it must be the weak acid.

$\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+$

Only a small fraction of the molecules dissociate, so the concentration of ions is low and the conductivity is poor.

B is blue → basic → it must be the weak base.

$\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$

Again only a small fraction reacts, so few ions are present and the conductivity is poor.

Step 4 — Check that both measurements were needed

	Identity	Conductivity tells us	Colour tells us
A	$\text{CH}_3\text{COOH}$	weakly dissociated	acidic
B	$\text{CH}_3\text{NH}_2$	weakly dissociated	basic
C	$\text{NH}_4\text{Cl}$	fully ionised	acidic

Neither measurement alone was sufficient. Conductivity separated the fully ionised salt from the two weak electrolytes but could not tell A from B. The indicator separated acidic from basic but could not tell A from C, since both are acidic.

Note especially that C is acidic AND a good conductor, while A is acidic and a poor one. That combination is the whole point of the question: conductivity measures ions, not acidity.