Solution Stoichiometry and Titration Calculations

Molarity is moles of solute per litre of solution, so it works in both directions. Multiply a volume in litres by molarity to get moles, or divide moles by molarity to get the volume needed. Treat it exactly as you treat molar mass in a mass-based problem.

The units supply a useful check: (mol/L) × L = mol. For 12.50 mL of 0.0800 M AgNO₃, n = 0.0800 × 0.01250 = 0.00100 mol Ag⁺. A burette volume is final reading minus initial reading; readings of 3.20 and 21.80 mL mean 18.60 mL delivered, not 21.80 mL. Read the same scale and precision at both endpoints.

Convert the given quantity to moles, apply the mole ratio from the balanced equation, then convert to whatever is requested. The only thing that changes between problems is which conversion sits at each end: molarity for solutions, molar mass for solids, and the ideal gas law for gases.

When both reactants have finite amounts, compare available moles divided by their coefficients before calculating yield. Mixing 20.0 mL of 0.100 M AgNO₃ with 30.0 mL of 0.0500 M NaCl supplies 0.00200 mol Ag⁺ and 0.00150 mol Cl⁻. Chloride limits the 1:1 reaction, giving 0.00150 mol AgCl and leaving 0.00050 mol Ag⁺. With additive volumes, the residual silver concentration is 0.00050/0.0500 = 0.0100 M. Use each original volume to count supplied moles, then the total volume for the final concentration. The precipitation yield treats the small equilibrium solubility as negligible.

A titrant of known concentration is delivered from a burette into a measured volume of analyte until the reaction is exactly complete. Recording the initial and final burette readings gives the volume delivered, which is the single measurement the whole calculation depends on.

A titration measures the analyte in the transferred aliquot. If 25.00 mL from a 250.0 mL diluted flask is titrated, it contains one tenth of the flask's analyte, assuming complete mixing. Multiply the aliquot amount by 250.0/25.00 before referring back to the original sample. Dilution conserves analyte moles; the subsequent reaction consumes them according to a balanced equation.

Rinse the burette with titrant so residual water does not dilute its contents. Rinse the transfer pipette with the analyte solution, but rinsing the titration flask with distilled water ordinarily changes neither analyte moles nor equivalence volume. Water in that flask changes concentrations, not the stoichiometric amount of titrant required.

The equivalence point is the theoretical moment when the moles added exactly satisfy the stoichiometry. The endpoint is when the indicator changes colour. A well-chosen indicator makes them nearly coincide, and the small gap between them is the main systematic error in the technique.

The indicator or instrumental signal is selected so its endpoint closely estimates equivalence. Overshooting with extra titrant makes the recorded volume too large; if you use that volume with the stated titrant concentration, you overestimate analyte moles. Repeat concordant trials and discard a rough scouting titration according to the lab protocol rather than averaging a clear overshoot into precise trials. Equivalence alone does not imply neutral pH; a strong acid–strong base mixture is neutral at equivalence in the ideal model, with pH 7 specifically at 25 °C.