How this calculation works
A very dilute weak acid cannot be modeled reliably by simply taking the square root of concentration times Ka. Water supplies hydrogen and hydroxide ions even when the analytical amount of acid approaches zero. This calculation solves the ideal concentration charge balance for one monoprotic acid or neutral base in otherwise pure water. The water term remains present throughout the calculation, including the neutral limiting case.
x − Kw/x = C K/(K+x); x is H+ for acid or OH− for base; Kw = 10⁻¹⁴A worked example
For a one-millimolar acid with Ka one times ten to the minus five, the main dissociation calculation gives pH about 4.02. This is a concentration-based educational result, not an instrument reading. At the same formal concentration and Kb, the base pH complements the acid pH to fourteen. Reducing formal concentration far enough makes either answer approach the neutral value instead of crossing to an impossible opposite-side dilute limit.
Good to know
Added salts, buffers and multiple acid sites require additional balances and are excluded. The page does not determine a titration curve or accept a mixture of strong and weak acids. Activity effects become important outside dilute ideal behavior. The broad numeric input range permits mathematical scenarios; it does not establish that concentration-based pH accurately represents a concentrated real solution.
Reference: OpenStax — weak acid and base equilibria (opens in a new tab) ↗
What this calculation answers
A very dilute weak acid cannot be modeled reliably by simply taking the square root of concentration times Ka. Water supplies hydrogen and hydroxide ions even when the analytical amount of acid approaches zero.
This calculation solves the ideal concentration charge balance for one monoprotic acid or neutral base in otherwise pure water. The water term remains present throughout the calculation, including the neutral limiting case.
The Strong acid base titration curve calculator tracks a strong-acid or strong-base stoichiometric titration. The present page instead solves one weak species in pure water, including water autoionization.
Use the right input conventions
Enter the total formal concentration before dissociation, in moles per litre. For acid mode the equilibrium constant is Ka; for base mode it is Kb, not the Ka of the conjugate acid.
The numerical model fixes temperature at twenty-five degrees Celsius and water ionic product at ten to the minus fourteen. Choosing base mode changes which ion is solved first, rather than changing that water constant.
Interpret the result
The output separates pH, hydrogen concentration and hydroxide concentration. Their product follows the fixed water ionic product, while the dissociated species fraction follows the entered equilibrium constant and solved primary ion concentration.
A zero formal concentration gives pH seven in this model, but its ionized fraction is undefined because there is no dissolved formal species to divide by. It is displayed as undefined rather than assigned a misleading zero percentage.
Work through the example
For a one-millimolar acid with Ka one times ten to the minus five, the main dissociation calculation gives pH about 4.02. This is a concentration-based educational result, not an instrument reading.
At the same formal concentration and Kb, the base pH complements the acid pH to fourteen. Reducing formal concentration far enough makes either answer approach the neutral value instead of crossing to an impossible opposite-side dilute limit.
Where the model stops
Added salts, buffers and multiple acid sites require additional balances and are excluded. The page does not determine a titration curve or accept a mixture of strong and weak acids.
Activity effects become important outside dilute ideal behavior. The broad numeric input range permits mathematical scenarios; it does not establish that concentration-based pH accurately represents a concentrated real solution.
A few questions, answered
Why retain the water term?
It supplies both ions even without added acid or base and preserves the neutral dilute limit.
What changes in the comparison?
A much smaller formal concentration brings the solution toward pH seven instead of following the square-root shortcut indefinitely.