Solution stoichiometry — titrations, dilutions

1. Concentration — what is molarity, really?

WHY this definition? Chemistry happens between particles. We can't scoop particles, but if we know concentration and pour a measured volume, we know exactly how many moles (hence particles) we handed over.

WHAT is $n$ ? Recall $n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{m}{M_r}$ .

2. Dilution — adding water to weaken a solution

3. Titrations — counting unknown moles

Figure — Solution stoichiometry — titrations, dilutions

Worked example Acid–base titration (1:1)

$25.0\text{ mL}$ of NaOH of unknown $c$ needs $22.5\text{ mL}$ of $0.100\text{ M}$ HCl to neutralise. Reaction: $\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$ (ratio 1:1).

$n_{\text{HCl}} = c V = 0.100 \times 0.0225 = 2.25\times10^{-3}\text{ mol}$ . Why litres now? Because we want moles, and $c$ is mol L⁻¹.
Ratio 1:1 → $n_{\text{NaOH}} = 2.25\times10^{-3}\text{ mol}$ . Why? One NaOH per HCl.
$c_{\text{NaOH}} = \dfrac{n}{V} = \dfrac{2.25\times10^{-3}}{0.0250} = 0.0900\text{ M}$ .

Worked example Non-1:1 ratio (the ratio really matters)

$20.0\text{ mL}$ of $\text{H}_2\text{SO}_4$ is neutralised by $30.0\text{ mL}$ of $0.200\text{ M}$ NaOH. Find $c_{\text{H}_2\text{SO}_4}$ . Reaction: $\text{H}_2\text{SO}_4 + 2\,\text{NaOH} \to \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ .

$n_{\text{NaOH}} = 0.200 \times 0.0300 = 6.00\times10^{-3}\text{ mol}$ .
$\dfrac{n_{A}}{1} = \dfrac{n_{\text{NaOH}}}{2}$ → $n_{\text{H}_2\text{SO}_4} = \tfrac{1}{2}(6.00\times10^{-3}) = 3.00\times10^{-3}\text{ mol}$ . Why halve? Each H₂SO₄ supplies 2 protons, so needs 2 NaOH.
$c = \dfrac{3.00\times10^{-3}}{0.0200} = 0.150\text{ M}$ . Forecast-then-verify: if you forgot the 2 you'd get $0.300$ M — double the truth. The ratio is not optional.

4. Back-titration (the 80/20 power move)

Flashcards

Define molarity (with units).

Moles of solute per litre of solution,

c=n/V

, unit mol L⁻¹ (M).

State the master relation linking moles, concentration, volume.

n = cV

Why does

c_1V_1=c_2V_2

hold for dilution?

Adding solvent changes volume but adds no solute, so moles

n=cV

are conserved.

In a titration, what bridges moles of A to moles of B?

The mole ratio from the balanced equation:

n_A/a = n_B/b

Give the general titration formula for

aA+bB

\dfrac{c_A V_A}{a} = \dfrac{c_B V_B}{b}

Difference between equivalence point and end point.

Equivalence point = stoichiometrically exact reaction; end point = observed indicator colour change (ideally coincide).

For

\text{H}_2\text{SO}_4 + 2\text{NaOH}

, how do moles relate?

n_{\text{NaOH}} = 2\,n_{\text{H}_2\text{SO}_4}

Which volume goes into

n=cV

in a titration?

The delivered/pipetted volume of that solution, NOT the mixed total.

Formula for moles reacted in a back-titration.

n_{\text{reacted}} = n_{\text{excess added}} - n_{\text{leftover titrated}}

Can you use mL on both sides of

c_1V_1=c_2V_2

Yes — units cancel as long as both sides use the same volume unit.

Recall Feynman: explain to a 12-year-old

Imagine orange squash. Concentration is how strong it tastes. Adding water (dilution) makes it weaker, but the amount of orange syrup inside is exactly the same — you just spread it in more water. That's why $c_1V_1=c_2V_2$ : syrup doesn't multiply. A titration is like this: you have mystery-strength lemon juice and you slowly add drops of baking-soda water until fizzing just stops. You know how strong your baking soda is and how many drops it took, so — using the recipe (balanced equation) that says "1 lemon needs 1 soda" — you work backwards to how strong the lemon was. The recipe is the magic bridge.

Connections

Mole concept and Avogadro's number — supplies $n = m/M_r$ and $n=cV$ .
Balancing chemical equations — the source of the all-important mole ratio.
Limiting reagent and excess — back-titration is a limiting-reagent problem in disguise.
Acid–base neutralisation — the most common titration reaction.
Indicators and pH — how the end point is actually detected.
Standard solutions and primary standards — how the "known" titrant is prepared.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, solution stoichiometry ka pura khel sirf ek line par tika hai: moles = concentration × volume, yaani $n = cV$ . Concentration (molarity) matlab ek litre solution mein kitne mole ghula hai. Jab tumhe particles count karne hote hain reaction ke liye, tum seedha particles to utha nahi sakte — isliye known concentration ka known volume daal do, aur moles pata chal jaate hain.

Dilution samajhna sabse easy hai. Jab tum paani milate ho, tum solute (asli cheez) bilkul add nahi kar rahe — sirf volume badha rahe ho. Isliye moles same rehte hain, bas concentration kam ho jaati hai. Yahi se aata hai $c_1V_1 = c_2V_2$ . Yaad rakho — dono taraf same unit (mL ya L) use karo, cancel ho jaayega.

Titration mein hum ek unknown solution ki strength nikalte hain known solution (titrant) se react kara ke, jab tak reaction exactly complete na ho jaaye (equivalence point, indicator se pata chalta hai). Logic chain yeh hai: known ka $cV$ se moles nikalo → balanced equation ka mole ratio cross karo → doosri cheez ke moles → uska $cV$ laga ke unknown nikaalo. Sabse badi galti students yeh karte hain ki mole ratio bhool jaate hain. Jaise $\text{H}_2\text{SO}_4 + 2\text{NaOH}$ mein, ek H₂SO₄ ko do NaOH chahiye — agar '2' bhool gaye to answer double galat aayega. Isliye pehle hamesha balanced equation likho, phir coefficients se divide karo.

Solution stoichiometry — titrations, dilutions

1. Concentration — what is molarity, really?

2. Dilution — adding water to weaken a solution

3. Titrations — counting unknown moles

4. Back-titration (the 80/20 power move)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Chemical Reactions & Stoichiometry

Connections