Acid-Base Titration Calculator

How this titration calculator works

An acid-base titration lets you determine the concentration of an unknown solution by reacting it with a second solution whose concentration is already known. In practice, the known solution is the titrant in the burette, and the unknown solution is the analyte in the flask. When enough titrant has been added to satisfy the balanced reaction, the measured endpoint volume can be used to calculate how much analyte was present. This calculator performs that stoichiometric step automatically and presents the result in a form that is easy to check against lab notes or homework.

The page is built for the most common classroom and routine laboratory use case: you know the titrant molarity, you know how much titrant was delivered, you know the volume of analyte that was titrated, and you know the mole ratio from the balanced equation. From those values, the calculator finds the analyte concentration. If you also enter molar mass, it estimates the mass of analyte contained in the aliquot you titrated. The curve display and mini-game are included as teaching aids so the endpoint is not just a number but something you can connect to the chemistry of the run.

The most common source of confusion in titration problems is not the algebra itself. It is usually one of three things: mixing up titrant and analyte, forgetting to convert milliliters to liters, or assuming every neutralization is 1:1. This page is meant to reduce those mistakes by labeling each input clearly and by keeping the explanation close to the calculator. If you already know the method, you can use the form immediately. If you are still learning, the sections below explain what each field means, how the formula is arranged, and how to interpret the result.

Inputs and what they mean

The first input is the titrant concentration, written as $C_t$, in moles per liter. This is the known molarity of the solution delivered from the burette. The second input is the titrant volume, $V_t$, in milliliters. Enter the volume actually delivered to reach the endpoint, which is usually the final burette reading minus the initial burette reading. The third input is the analyte volume, $V_a$, also in milliliters. This is the sample volume of the unknown solution that was placed in the flask and titrated.

The stoichiometric ratio is the number of moles of titrant that react with one mole of analyte according to the balanced chemical equation. For a simple strong acid-strong base neutralization such as HCl with NaOH, the ratio is often 1. For a diprotic acid titrated by a strong base, the ratio may be 2 because two moles of hydroxide are required for each mole of acid. The calculator does not guess the chemistry from compound names, so this field matters. If the ratio is wrong, the concentration result will also be wrong even if the measured volumes are perfect.

The molar mass field is optional. It does not change the concentration calculation. Instead, it converts the calculated moles of analyte into grams. That is useful when a lab report asks for the amount of substance in the aliquot as both moles and mass. If you only need molarity, you can leave molar mass blank and the calculator will still return the concentration and moles of analyte.

Core formula

At the equivalence point, the reacting amounts are linked by stoichiometry. The calculator uses the relationship:

$C_t{V}_t=r{C}_a{V}_a$

Solving for the unknown analyte concentration gives:

$C_a=\frac{C_t{V}_t}{r{V}_a}$

Here, $C_t$ is the titrant concentration, $V_t$ is the titrant volume at the endpoint, $V_a$ is the analyte volume, $r$ is the stoichiometric ratio, and $C_a$ is the unknown analyte concentration. The form accepts volumes in milliliters because that is how burettes and pipettes are usually read, but the script converts those values to liters internally before applying molarity relationships.

Once the titrant moles are known from $n_t=C_t{V}_t$, the moles of analyte follow from the reaction ratio. Dividing analyte moles by analyte volume gives the analyte concentration. If molar mass is supplied, mass is then found from moles multiplied by grams per mole. This is standard analytical chemistry, but the calculator keeps the steps organized so you can focus on the experiment rather than repetitive arithmetic.

Worked example

Suppose 25 mL of an unknown monoprotic acid is titrated with 0.100 M sodium hydroxide, and the endpoint is reached after 30.0 mL of base has been delivered. Because one mole of NaOH reacts with one mole of the acid, the stoichiometric ratio is 1. Enter 0.100 for titrant concentration, 30.0 for titrant volume, 25.0 for analyte volume, and 1 for the ratio.

The concentration is:

$C_a=\frac{0.1\times 30}{1\times 25}=0.12M$

So the unknown acid concentration is 0.12 M. The moles of acid in the 25 mL aliquot are 0.12 × 0.025 = 0.003 mol. If the analyte molar mass were 98.08 g/mol, the mass in that aliquot would be about 0.294 g. This example shows the main idea of titration clearly: a measured endpoint volume, combined with known stoichiometry, reveals the amount of unknown substance in the sample.

How to interpret the result

The concentration shown in the result box is the concentration of the analyte solution that you titrated, not the concentration of the titrant. The moles shown are the moles present in the analyte sample volume you entered. If you provide molar mass, the mass shown is the mass of analyte in that same aliquot. If your aliquot came from a larger diluted flask or from a stock solution, you may need an additional dilution calculation outside this tool to relate the result back to the original sample.

It is also important to remember that the calculator assumes the endpoint volume entered is the correct delivered volume for the analysis. In a real experiment, the observed endpoint may differ slightly from the true equivalence point because of indicator range, overshoot, or reading uncertainty. The stoichiometric calculation is still the right starting point, but your final reported answer should reflect the quality of the measurements used to produce it.

Curve and assumptions

The titration curve visualization below the form is a simplified educational model. It helps you see that pH changes gradually at first and then much more sharply near equivalence in a strong acid-strong base style titration. That steep region is why careful dropwise addition matters near the endpoint. A very small extra volume can shift the pH quickly and cause an overshoot. The graph is useful for intuition, but it is not a full equilibrium solver for every weak acid, weak base, buffer, or polyprotic system.

The main concentration calculation remains broadly useful as long as the balanced reaction and stoichiometric ratio are correct. Still, no calculator can fix poor experimental technique. Burette reading errors, air bubbles, contamination, missed dilution factors, or incorrect reaction balancing will all affect the answer. The best way to use this page is as a reliable stoichiometric calculator paired with careful laboratory practice and sensible interpretation of the data.

Reference formulas and relationships

The original calculator formulas are preserved below in MathML so the page remains machine-readable, accessible to supporting tools, and faithful to the chemistry relationships used by the script.

$C_t=\frac{n_t}{V_t}$

$C_a=\frac{n_a}{V_a}$

$n_t=C_t{V}_t$

$n_a=\frac{n_t}{r}$

$n_a=C_a{V}_a$

$m=n_{a}M$

$V_t\left(L\right)=\frac{V_t\left(\mathrm{mL}\right)}{1000}$

$V_a\left(L\right)=\frac{V_a\left(\mathrm{mL}\right)}{1000}$

$C_t{V}_t=r{C}_a{V}_a$

$C_a=\frac{C_t{V}_t}{r{V}_a}$

$r=\frac{n_t}{n_a}$

$n_t=r{n}_a$

$n_a=\frac{C_t{V}_t}{r}$

$C_a{V}_a=\frac{C_t{V}_t}{r}$

$m=\frac{C_t{V}_{t}M}{r}$

$\mathrm{pH}=-{\log }_{10}\left(\right[H^{+}\left]\right)$

$\mathrm{pOH}=-{\log }_{10}\left(\right[{\mathrm{OH}}^{-}\left]\right)$

$\mathrm{pH}+\mathrm{pOH}=14$