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1.3.5 · D5Chemical Reactions & Stoichiometry

Question bank — Solution stoichiometry — titrations, dilutions

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Figure — Solution stoichiometry — titrations, dilutions

True or false — justify

Molarity uses the volume of solvent you added.
False — it uses the total volume of the solution after topping up. The dissolved solute takes up space too, so solvent volume ≠ solution volume. See the parent's volumetric-flask fix.
Diluting a solution changes the number of moles of solute.
False — pure water carries zero moles of solute, so stays fixed while grows and falls. That conservation is exactly why .
For the relation is the same equation as the dilution law .
False — they are algebraically identical but conceptually opposite: dilution conserves one substance's moles across a volume change, while titration equates moles of two different reacting substances via a ratio.
In you must convert both volumes to litres.
False — but with a catch: the concentrations must be in the same unit, and the volumes in the same unit as each other; then the volume unit cancels and mL/mL is fine. It only stays safe because no lone multiplies a concentration to make moles here.
At the equivalence point the indicator changes colour exactly.
Not necessarily — the equivalence point is where moles are stoichiometrically exact; the end point is where the indicator flips. A good indicator makes them nearly coincide, but they are different events. Compare Indicators and pH.
Adding more water to a titration flask changes how much titrant you need.
False — extra water changes the mixture's concentration but not the moles of analyte present, and titrations count moles. The required titre depends only on moles delivered, not on dilution of the flask.
For , equal volumes of equal-concentration solutions will neutralise exactly.
False — the acid supplies two protons per molecule, so it needs twice the moles of base; you'd need double the volume (or double the concentration) of NaOH. This is a limiting-reagent idea — the base runs out first if under-supplied; see Limiting reagent and excess.
A back-titration measures the reagent that reacted with the analyte directly.
False — it measures the leftover excess reagent; the reacted amount is inferred as .

Spot the error

"I dissolved 1 mol NaOH in 1 L of water, so my solution is 1.0 M."
The final solution volume exceeds 1 L once the solid dissolves, so the concentration is slightly less than 1.0 M. Dissolve, then top up to the 1 L mark.
" neutralised by NaOH, so gives the answer."
The ratio was ignored; the correct bridge is , i.e. . Always balance first — see Balancing chemical equations.
"After mixing, the flask holds 47.5 mL, so I plug 47.5 mL into ."
You must use the volume of the one solution whose moles you want (the pipetted analyte or the burette titre), not the combined mixture volume. Moles, not mixture concentration, are tracked.
"Molarity is mass over volume."
It is moles over volume, . Convert mass to moles first via using the molar mass. Revisit Mole concept and Avogadro's number.
"I got mol HCl using with mL."
Volume must be in litres when it multiplies a mol L⁻¹ concentration: . Using 22.5 inflates the answer by 1000×.
"Since was measured, the analyte reacted with nothing."
If leftover equals added, then — meaning your "excess" wasn't actually consumed at all, signalling a failed reaction or wrong setup, not a valid result.
"I standardised my HCl against ordinary bench NaOH pellets."
NaOH is hygroscopic and absorbs CO₂, so it is not a primary standard; its true concentration is uncertain. Use a proper primary standard instead — see Standard solutions and primary standards.

Why questions

Let and be the stoichiometric coefficients in . Why divide by them in ?
Because moles of A react with exactly moles of B; dividing each mole count by its own coefficient normalises both to the same "per reaction unit," which makes the two sides equal.
Why can't you skip the balanced equation in a titration?
The balanced equation supplies the mole ratio (the coefficients ) — the only bridge from moles of the known substance to moles of the unknown. Without it you have no way to connect the two amounts. See Acid–base neutralisation.
Why does back-titration exist if direct titration is simpler?
It handles analytes that are insoluble, slow to react, or lack a sharp end point — you react them with known excess reagent, then titrate what's left over.
Why is called the "master relation"?
Every formula in the topic — dilution (), the general titration law, and back-titration bookkeeping — is just applied and conserved in different situations.
Why must the titrant be a solution of accurately known concentration?
The whole calculation works backwards from the titrant's known moles; any error in its concentration propagates directly into the unknown answer. Hence standard solutions — see Standard solutions and primary standards.

Edge cases

If you accidentally overshoot the end point by one drop, is the equivalence point moles equation still exactly true?
No — you've added slightly more titrant than stoichiometry demands, so slightly exceeds the exact value and the computed unknown is a touch too high.
What does predict if you dilute a solution to infinite volume?
As , . The moles are still there but so spread out that concentration approaches zero — you can dilute weaker forever but never remove the solute.
If you add pure solute (not water) to a solution, does still hold?
No — that law assumes is conserved. Adding solute changes , so you must recompute total moles, not apply the dilution law.
In a titration, if exactly, what does that tell you?
With a ratio and equal volumes, — the two solutions had identical concentrations. It's the only case where equal titre means equal strength.
What happens to the titration logic if the reaction has a coefficient (a spectator not in the equation)?
Spectator ions don't appear in the balanced net reaction, so they carry no mole ratio and are simply ignored — only the reacting species set the bridge.
If a diprotic acid is only partially neutralised (one proton), what ratio applies?
Then effectively for that stage, not — the ratio depends on how many protons the reaction condition actually removes, so the reaction you write must match the observed end point.
Recall The one habit that beats every trap

Before any titration calculation, run these four steps in order:

  1. Write the balanced equation so the coefficients are visible.
  2. Box the mole ratio — that is your bridge.
  3. Convert every volume to litres before it multiplies a concentration in (leave volumes alone inside ).
  4. Ask "moles of which substance?" — pick the substance whose and are both known, compute its , then cross the ratio bridge to the unknown. When several species are present, the "which" is decided by the reaction you wrote: only species that appear in the balanced equation carry moles that matter; spectators are ignored.

Connections