Chemistry · General chemistry II · Worked example
Find an unknown concentration by titration
A 25.00 mL sample of hydrochloric acid is titrated with 0.1000 M NaOH, and the equivalence point is reached after 18.50 mL. Find the concentration of the acid.
Moles of titrant
Convert the buret volume to liters and multiply by the molarity.
Moles of analyte
HCl + NaOH → NaCl + H₂O: one mole of acid for each mole of base, so 1.850 × 10⁻³ mol of HCl reacted.
Concentration
Divide by the acid’s own volume, 25.00 mL, not the combined volume in the flask.
Count the significant figures
Every measurement has four significant figures, so the answer keeps four: 0.07400 M, trailing zeros included.
Result
The hydrochloric acid is 0.07400 M.
Your turn
A 10.00 mL sample of vinegar needs 16.20 mL of 0.5000 M NaOH to reach the equivalence point. Acetic acid reacts 1 : 1 with NaOH. Find its concentration.
Show the answer and explanation
0.8100 M.
n = (0.01620 L)(0.5000 mol/L) = 8.100 × 10⁻³ mol of NaOH, so 8.100 × 10⁻³ mol of acetic acid, and 8.100 × 10⁻³ mol ÷ 0.01000 L = 0.8100 M. A weak acid still reacts completely with a strong base, so the stoichiometry is the same.
Keep exploring
Buffers & titration curves opens with 25.00 mL of 0.07400 M HCl and 0.1000 M NaOH. Its equivalence point falls at 18.50 mL, where the pH is 7.00.
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