Library / Solution chemistry, equilibria and kinetics / SCH-05
Solution chemistry, equilibria and kinetics
Missing ion concentration from charge balance
Concentration of one unlisted ion that makes the solution electrically neutral, from the signed charge numbers z and concentrations c of up to three other ions.
- Ref
- SCH-05
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- cx [mol/L]
A solution is electrically neutral, so the charges of the cations and anions add up to zero: Σ zᵢcᵢ + zₓcₓ = 0. The concentration of one missing species follows as cₓ = −Σ zᵢcᵢ/zₓ. A negative answer means the inputs are inconsistent or another ion has been omitted. Use signed charge numbers, positive for cations.
Concentration of one unlisted ion that makes the solution electrically neutral, from the signed charge numbers z and concentrations c of up to three other ions. A negative result means the inputs are inconsistent or a species is missing.
Calculator
- Net charge of the listed ions, Σ z c0.04 eq/L
- Concentration of the listed ions, Σ c0.03 mol/L
- Charge carried by the missing ion, zₓ cₓ−0.04 eq/L
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.
- The listed ions carry 0.04 eq/L of charge, so 0.04 mol/L of an ion of charge −1 is needed to balance it.
- A result that is small or negative against the other concentrations points to an omitted species, or to an analytical error of that size.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/L) | Local step | Change in result |
|---|---|---|---|---|---|---|
| c2 Concentration of ion 2 | 0 to 1 mol/L | 0.02 | 2.02 | 2 | +10% | +5 % |
| c1 Concentration of ion 1 | 0 to 1 mol/L | 0.02 | 1.02 | 1 | +10% | +5 % |
| c3 Concentration of ion 3 (zero if unused) | 0 to 1 mol/L | 0.04 | 1.04 | 1 | +10% | 0 % |
| z1 Charge number of ion 1 | −4 to 4 | −0.06 | 0.1 | 0.16 | +1 | +50 % |
| z2 Charge number of ion 2 | −4 to 4 | −0.02 | 0.06 | 0.08 | +1 | +25 % |
| zx Charge number of the missing ion (not zero) | −4 to −1 | 0.01 | 0.04 | 0.03 | +1 | — |
| z3 Charge number of ion 3 | −4 to 4 | 0.04 | 0.04 | 0 | +1 | 0 % |
Sweep of Concentration of ion 2
| Concentration of ion 2 (mol/L) | Missing ion concentration (mol/L) |
|---|---|
| 0 | 0.02 |
| 0.125 | 0.27 |
| 0.25 | 0.52 |
| 0.375 | 0.77 |
| 0.5 | 1.02 |
| 0.625 | 1.27 |
| 0.75 | 1.52 |
| 0.875 | 1.77 |
| 1 | 2.02 |
Sweep one variable
Variables and how each one contributes
Missing ion concentration
What the equation gives you. Everything else feeds it.
Charge number of ion 1
The signed charge number of ion 1. Changes the charge it contributes in proportion. Raising it by 10 % (to 1.1) moves the result by +5 %, from 0.04 to 0.042 mol/L.
Concentration of ion 1
The concentration of ion 1, in mol/L. Changes the charge it contributes in proportion. Raising it by 10 % (to 0.022) moves the result by +5 %, from 0.04 to 0.042 mol/L.
Charge number of ion 2
The signed charge number of ion 2. Changes the charge it contributes in proportion. Raising it by 10 % (to 2.2) moves the result by +5 %, from 0.04 to 0.042 mol/L.
Concentration of ion 2
The concentration of ion 2, in mol/L. Changes the charge it contributes in proportion. Raising it by 10 % (to 0.011) moves the result by +5 %, from 0.04 to 0.042 mol/L.
Charge number of ion 3
The signed charge number of ion 3. Changes the charge it contributes in proportion (matters only if c₃ is not zero). Raising it by 10 % (to 1.1) moves the result by +0 %, from 0.04 to 0.04 mol/L.
Concentration of ion 3 (zero if unused)
The concentration of ion 3, in mol/L (zero if unused). Changes the charge it contributes in proportion. Raising it by 10 % (to 0) moves the result by +0 %, from 0.04 to 0.04 mol/L.
Charge number of the missing ion (not zero)
The signed charge number of the missing ion, not zero. A more highly charged ion balances the same charge with a lower concentration, in inverse proportion. Raising it by 10 % (to −1.1) moves the result by −9.09 %, from 0.04 to 0.03636 mol/L.
Worked example with the values above
- Charge of the listed ions: 1×0.02 + 2×0.01 + 1×0 = 0.04 eq/L.
- The missing ion must carry the opposite charge: −0.04.
- cₓ = −0.04 / −1 = 0.04 mol/L.
Assumptions and limits
- Ideal dilute solutions, with concentrations in mol/L and masses in grams unless stated.
- Complete mixing and no evaporation or contraction on mixing unless the inputs say otherwise.
Common mistakes
- Mixing molality (per kg of solvent) with molarity (per litre of solution).
- Entering a mass percentage where a fraction between 0 and 1 is needed.
- Leaving out counterions when adding up charges.
Statistics and uncertainty
Masses and volumes are usually known to 0.1 % to 1 %, and concentrations from them carry the same relative error. Volumetric glassware tolerances add up through serial dilutions, so each extra stage widens the uncertainty.
Where it is used
- Preparing standards, stocks and dilution series in the laboratory.
- Checking charge balance and ionic strength in water analysis.
Maths you need
- Amount of substance: n = m/M, in mol
- Concentration: c = n/V, in mol/L
- Electroneutrality: total positive charge equals total negative charge