What does my titration give?
Type the titrant molarity, the volume you used to reach the end point, and the sample volume. The titration calculator gives the analyte’s molarity, using the mole ratio from your balanced equation, or the titrant volume you will need.
- Answer
- 0.17615 mol/L
The answer is 0.17615 mol/L.
- Analyte molarity (mol/L)
- 0.17615
- Titrant volume (mL)
- 35.23
- Titrant (mol)
- 0.0088075
- Analyte (mol)
- 0.0088075
Answer: 0.17615 mol/L. The answer is 0.17615 mol/L.
How to calculate
Works out the unknown concentration of an analyte from a titration (titrant molarity, titrant volume at the end point, sample volume and the mole ratio), or the titrant volume needed to reach the equivalence point.
Example with the default inputs (Find Analyte molarity, Titrant molarity, mol/L 0.25, Titrant volume used 35.23 mL, Analyte volume 50 mL, Analyte coefficient 1, Titrant coefficient 1): The answer is 0.17615 mol/L.
Method: n_titrant = C_titrant × V_titrant; n_analyte = n_titrant × a ÷ t; C_analyte = n_analyte ÷ V_analyte. For the volume: V_titrant = C_analyte × V_analyte × t ÷ (a × C_titrant).
- The end point you saw is the equivalence point, where analyte and titrant have reacted exactly in the ratio of the balanced equation.
- a and t are the coefficients of the analyte and the titrant in the balanced equation.
Worked examples
Each example is checked against the calculator on every build.
- Find Analyte molarity, Titrant molarity, mol/L 0.25, Titrant volume used 35.23 mL, Analyte volume 50 mL, Analyte coefficient 1, Titrant coefficient 1 gives Analyte molarity, mol/L 0.17615, Answer 0.17615 mol/L.Source: OpenStax, Chemistry 2e, §4.5 Quantitative Chemical Analysis (titration: moles from molarity × volume and the mole ratio of the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-5-quantitative-chemical-analysis, Example 4.14 Titration Analysis: 0.176 M HCl
- Find Analyte molarity, Titrant molarity, mol/L 0.09113, Titrant volume used 23.24 mL, Analyte volume 20 mL, Analyte coefficient 5, Titrant coefficient 2 gives Analyte molarity, mol/L 0.264733.Source: OpenStax, Chemistry 2e, §4.5 Quantitative Chemical Analysis (titration: moles from molarity × volume and the mole ratio of the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-5-quantitative-chemical-analysis, Check Your Learning after Example 4.14 (the book prints 0.2648 M)
- Find Titrant volume, Titrant molarity, mol/L 0.1, Analyte volume 25 mL, Analyte molarity, mol/L 0.1, Analyte coefficient 1, Titrant coefficient 2 gives Titrant volume (mL) 50, Analyte (mol) 0.0025.Source: OpenStax, Chemistry 2e, §4.5 Quantitative Chemical Analysis (titration: moles from molarity × volume and the mole ratio of the balanced equation), https://openstax.org/books/chemistry-2e/pages/4-5-quantitative-chemical-analysis
How it works
Write the balanced equation as a·analyte + t·titrant → products. At the equivalence point:
- moles of titrant n_t = C_t × V_t (V in liters)
- moles of analyte n_a = n_t × a ÷ t
- analyte molarity C_a = n_a ÷ V_a
To plan a titration, the titrant volume is V_t = C_a × V_a × t ÷ (a × C_t). The page shows it in mL.
The coefficients a and t are whole numbers from 1 to 100 (both 1 by default). Volumes: 1 mL = 0.001 L; 1 US fl oz = 3.785411784 L ÷ 128 = 0.0295735295625 L.
Exact arithmetic. Each value is read as the exact decimal you typed, in its unit, so each result is an exact fraction until it is rounded for display.
Output format. The answer line shows … mol/L or … mL of titrant, the number to 6 significant figures with no thousands separators. The rows show the analyte molarity, the titrant volume in mL, and the moles of titrant and analyte, to 6 significant figures; values of 10¹⁵ or more, or below 10⁻⁶, show in scientific form.
When there is no answer. A planned titrant volume over 1,000 m³, or a value too small to hold as a number.
Assumptions
- The end point matches the equivalence point.
- Molarity is moles per liter of solution.
Worked examples by hand
HCl with NaOH (OpenStax Example 4.14). n(NaOH) = 0.250 × 0.03523 = 0.0088075 mol = n(HCl); C = 0.0088075 ÷ 0.05000 = 0.17615 mol/L (the book: 0.176 M).
Oxalic acid with permanganate (OpenStax Check Your Learning). n(MnO₄⁻) = 0.09113 × 0.02324 = 0.0021178612 mol; n(H₂C₂O₄) = × 5/2 = 0.005294653 mol; C = ÷ 0.02000 = 0.26473265 mol/L. The book prints 0.2648 M; the exact arithmetic gives 0.2647.
Planning a sulfuric acid titration. n(H₂SO₄) = 0.100 × 0.02500 = 0.0025 mol; n(NaOH) = 0.0025 × 2 = 0.0050 mol; V = 0.0050 ÷ 0.100 = 0.0500 L = 50 mL.
Other questions people ask
How do I calculate concentration from a titration?
Moles of titrant = molarity × liters used. Convert to moles of analyte with the mole ratio from the balanced equation, then divide by the sample volume in liters. For HCl and NaOH (1:1), 35.23 mL of 0.250 M NaOH in a 50.00 mL sample gives 0.176 M HCl.
What is the equivalence point?
The point where the titrant added has reacted exactly with all the analyte, in the ratio of the balanced equation. An indicator color change or a pH jump (the end point) shows when you reach it.
What if the mole ratio is not 1:1?
Set the coefficients under More options. For H₂SO₄ + 2NaOH, the acid’s coefficient is 1 and the base’s is 2, so each mole of NaOH neutralizes half a mole of acid.
Can I use M₁V₁ = M₂V₂ for a titration?
Only when the mole ratio is 1:1. In general, a × C_titrant × V_titrant = t × C_analyte × V_analyte, where a and t are the coefficients of the analyte and the titrant.
How much titrant will I need?
Pick Titrant volume and type your expected analyte molarity. 25.00 mL of 0.100 M H₂SO₄ needs 50.0 mL of 0.100 M NaOH, because each mole of acid takes two moles of base.
Does the volume unit matter?
No. Type mL, L or fluid ounces; the page converts to liters. Molarity is moles per liter of solution.