Titration
Inputs
| Solve for | Find acid concentration |
|---|---|
| Acid Concentration | 0.1 M |
| Acid Volume | 25 mL |
| Base Concentration | 0.1 M |
| Base Volume | 25 mL |
| Acid Equivalents | 1 |
| Base Equivalents | 1 |
Titration
Solve acid–base titration problems at the equivalence point using C(acid)·V(acid)·n(acid) = C(base)·V(base)·n(base). Find the unknown concentration or volume of either the acid or the base.
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Details
Acid–base titration
Titration is a technique for finding an unknown concentration by reacting the unknown solution with a standard solution of known concentration. The reaction is carried out dropwise until the equivalence point is reached — the moment when the acid and base have reacted in exactly the right stoichiometric ratio. At that point, a simple relationship connects the four quantities: concentration and volume of the acid, and concentration and volume of the base.
The titration formula
At the equivalence point, moles of acid equivalents equal moles of base equivalents:
Ca⋅Va⋅na=Cb⋅Vb⋅nb| Symbol | Meaning |
|---|---|
| Acid concentration (mol/L) | |
| Acid volume (mL or L) | |
| Acid n-factor (protons per formula unit) | |
| Base concentration (mol/L) | |
| Base volume (mL or L) | |
| Base n-factor (hydroxides per formula unit) |
Rearranging gives any one of the four main quantities when the other three are known. For example, to find the acid concentration:
Ca=Va⋅naCb⋅Vb⋅nbThe n-factor
The n-factor counts how many protons an acid donates (or how many hydroxide equivalents a base provides) per formula unit. For most common acids and bases it equals 1:
- n = 1: HCl, HNO₃, acetic acid (CH₃COOH), NaOH, KOH
- n = 2: H₂SO₄, H₂CO₃, Ca(OH)₂, Ba(OH)₂
- n = 3: H₃PO₄
When the formula reduces to the familiar . Using the wrong n-factor shifts the calculated concentration by a whole-number multiple, so it is worth confirming the acid and base identities before entering values.
Worked example
A student titrates 25.00 mL of an unknown hydrochloric acid solution against 0.200 M NaOH. The equivalence point is reached after adding 18.75 mL of base. Both are monoprotic ().
Ca=Va⋅naCb⋅Vb⋅nb=25.00 mL×10.200 M×18.75 mL×1=0.150 MThe moles of equivalents transferred at the equivalence point:
neq=Cb⋅Vb⋅nb=0.200 M×18.75 mL=3.75 mmolDiprotic acid example
Sulfuric acid (H₂SO₄) has . To titrate 10.00 mL of a 0.100 M H₂SO₄ solution against 0.200 M NaOH:
Vb=Cb⋅nbCa⋅Va⋅na=0.200 M×10.100 M×10.00 mL×2=10.00 mLThe two-fold n-factor exactly doubles the base volume needed compared to an equivalent monoprotic acid.
Equivalence point vs. endpoint
The equivalence point is a theoretical concept — the exact moment stoichiometric neutralisation is complete. The endpoint is the observable signal used in practice, typically a colour change of an indicator (phenolphthalein turns pink near pH 9, methyl orange turns yellow near pH 4). A well-chosen indicator places the endpoint within a fraction of a drop of the true equivalence point.
Back titration
When the analyte dissolves slowly or cannot be titrated directly, a back titration is used: an excess of standard reagent is added to the analyte, the reaction is allowed to go to completion, and the leftover reagent is then titrated with a second standard solution. The same equivalence-point formula applies twice, once for each titration step.
Frequently Asked Questions (FAQ)
What is the equivalence point in a titration?
The equivalence point is the moment during a titration when the acid and base have reacted in exactly the stoichiometric ratio — that is, all the acid equivalents have been neutralised by the base equivalents and no excess remains. At this point the formula C(acid)·V(acid)·n(acid) = C(base)·V(base)·n(base) holds exactly. The equivalence point is not the same as the endpoint: the endpoint is the colour change of an indicator, which ideally coincides with the equivalence point but may be slightly before or after.
What formula does this use?
At the equivalence point the moles of acid equivalents equal the moles of base equivalents:
C(acid) × V(acid) × n(acid) = C(base) × V(base) × n(base)
where C is the molar concentration, V is the volume, and n is the n-factors (number of acidic protons or hydroxide ions per formula unit). Rearranging gives any one unknown when the other three are known.
What is the n-factor and when is it not 1?
The n-factor (or equivalents factor) counts how many protons one molecule of acid donates, or how many hydroxides one molecule of base provides. For monoprotic acids (HCl, HNO₃, acetic acid) and monobasic bases (NaOH, KOH) the n-factor is 1 and the formula reduces to the familiar C₁V₁ = C₂V₂. For diprotic acids it is 2 (H₂SO₄ → 2 H⁺; H₂CO₃ → 2 H⁺) and for triprotic acids it is 3 (H₃PO₄ → 3 H⁺). Ca(OH)₂ has an n-factor of 2. Using the wrong n-factor will give a concentration that is off by a whole-number factor.
How does titration relate to pH?
Titration is about stoichiometry — balancing moles of acid and base — while pH measures the hydrogen-ion concentration in the solution. At the equivalence point of a strong acid–strong base titration the pH is 7; for a weak acid–strong base titration it is above 7 because the conjugate base hydrolyses. The titration formula here applies regardless of pH, because it only uses the amounts that react, not the strength of the acid or base. To trace the full pH curve through the titration, additional calculations involving Ka or Kb are needed.
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