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Library / Solution chemistry, equilibria and kinetics / SCH-22

Solution chemistry, equilibria and kinetics

Mixed hydroxide and carbonate from two endpoints

Amounts of hydroxide and carbonate in an aliquot titrated with strong monoprotic acid, from the cumulative volumes V₁ and V₂ to the first and second endpoints (phenolphthalein and methyl orange).

Ref
SCH-22
Rev
v1
Field
Solution chemistry, equilibria and kinetics
Result
nOH [mol]

A mixture of hydroxide and carbonate titrated with strong acid has two endpoints. Hydroxide is neutralised and carbonate goes to hydrogencarbonate by the first endpoint (V₁), and hydrogencarbonate is neutralised by the second (V₂). So the carbonate amount is c(V₂ − V₁) and the hydroxide amount is c(2V₁ − V₂). Both amounts must be non-negative, so V₁ ≤ V₂ ≤ 2V₁, and V₁, V₂ are cumulative volumes.

Amounts of hydroxide and carbonate in an aliquot titrated with strong monoprotic acid, from the cumulative volumes V₁ and V₂ to the first and second endpoints (phenolphthalein and methyl orange). Needs V₁ ≤ V₂ ≤ 2V₁. The carbonate amount is in the derived rows.

Calculator

Solve for
mol
mol/L
0.011
L
0.030.04
L
0.040.06
  • Amount of carbonate, c(V₂ − V₁)0.002 mol
  • Acid used to the first endpoint, c V₁0.003 mol
  • Acid used to the second endpoint, c V₂0.005 mol

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Amount of hydroxide (mol) current value 1e-3 mol
cAcid concentrationKey driver
1e-40.01
V1Cumulative volume to the first endpoint
0.003
V2Cumulative volume to the second endpoint
00.002
  • The aliquot contained 1×10−3 mol of hydroxide and 0.002 mol of carbonate, which used 0.003 mol of acid to the first endpoint.
  • Check: hydroxide plus twice the carbonate equals c V₂ = 0.005 mol, the total acid to the second endpoint.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mol)Local stepChange in result
c Acid concentration0.01 to 1 mol/L1e-40.010.0099+10%+10 %
V1 Cumulative volume to the first endpoint0.03 to 0.04 L1e-30.0030.002+10%+60 %
V2 Cumulative volume to the second endpoint0.04 to 0.06 L00.0020.002+10%−50 %

Sweep of Cumulative volume to the second endpoint

Cumulative volume to the second endpoint (L)Amount of hydroxide (mol)
0.040.002
0.04250.00175
0.0450.0015
0.04750.00125
0.051e-3
0.05257.5e-4
0.0555e-4
0.05752.5e-4
0.060

Sweep one variable

Amount of hydroxide (mol)Hover, touch, or use the arrow keys on the chart
05e−40.0010.00150.002 0.040.0450.050.0550.06 Cumulative volume to the second endpoint (L) now 1e-3

Variables and how each one contributes

nOH

Amount of hydroxide

Result · mol

What the equation gives you. Everything else feeds it.

c

Acid concentration

mol/L · typical 0.01 to 1

The acid concentration, in mol/L. Raises the hydroxide amount in direct proportion. Raising it by 10 % (to 0.11) moves the result by +10 %, from 1×10−3 to 0.0011 mol.

V1

Cumulative volume to the first endpoint

L · typical 0.03 to 0.04

The cumulative volume to the first endpoint, in L. Raises the hydroxide amount by 2c per litre. Raising it by 10 % (to 0.033) moves the result by +60 %, from 1×10−3 to 0.0016 mol.

V2

Cumulative volume to the second endpoint

L · typical 0.04 to 0.06

The cumulative volume to the second endpoint, in L. Lowers the hydroxide amount, as it raises the carbonate amount. Raising it by 10 % (to 0.055) moves the result by −50 %, from 1×10−3 to 5×10−4 mol.

Worked example with the values above

  1. Hydroxide term: 2V₁ − V₂ = 0.06 − 0.05 = 0.01 L.
  2. Multiply by c = 0.1 mol/L.
  3. nOH = 1×10−3 mol; nCO₃ = c(V₂ − V₁) = 0.002 mol.

Assumptions and limits

  • Complete, fast and well-defined reactions with a clear stoichiometry.
  • Volumes in litres or in a single consistent unit, and calibrated glassware and electrodes.

Common mistakes

  • Using volumes that are not cumulative.
  • Using V₂ below V₁ or above 2V₁, which gives negative amounts.
  • Forgetting that the second endpoint also neutralises the carbonate-derived hydrogencarbonate.

Statistics and uncertainty

Burette readings are good to ±0.02 to 0.05 mL, so a 25 mL titre has a relative error near 0.1 to 0.2 %, and the endpoint-detection error adds to it.

Where it is used

  • Analysing antacids, alkalinity, chloride and other titratable species.
  • Estimating the endpoint error of a titration.

Maths you need

  • Stoichiometry: mole ratios in a balanced equation
  • Equivalence point: where the reagent amounts match exactly
  • Titre: the volume of titrant delivered

Related formulas

Practise this formula