By Jack Brubaker
Updated Aug 30, 2022
Before starting, identify the titrant and analyte, then write the balanced reaction they undergo. For example, titrating acetic acid (CH3COOH) with sodium hydroxide (NaOH) yields:
CH3COOH + NaOH → CH3COONa + H2O
The coefficients on the left side give the molar ratio—in this case 1:1.
Record the burette reading at the end point (when the analyte is fully consumed). Convert milliliters to liters by dividing by 1,000. For instance, 39.75 mL of NaOH becomes:
39.75 mL ÷ 1,000 mL L-1 = 0.03975 L NaOH
Multiply the volume of titrant by its known concentration. If the NaOH solution is 0.1044 mol L-1:
0.03975 L × 0.1044 mol L-1 = 0.004150 mol NaOH
Use the molar ratio from Step 1 to convert titrant moles to analyte moles:
0.004150 mol NaOH × (1 mol CH3COOH ÷ 1 mol NaOH) = 0.004150 mol CH3COOH
Divide the moles of analyte by the volume of the sample (in liters). If 5.00 mL of vinegar was used:
5.00 mL = 0.00500 L
(0.004150 mol CH3COOH) ÷ 0.00500 L = 0.830 mol L-1
Thus, the vinegar contains 0.830 mol L-1 of acetic acid.