The standard calculation
2 marks · NormalA student titrates 25.0 cm³ of potassium hydroxide solution of unknown concentration with 0.150 mol/dm³ hydrochloric acid. The average volume of acid required to neutralise the alkali is 18.0 cm³. Calculate the number of moles of hydrochloric acid ( ) used in the reaction.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
0.00270 mol
Also accept: 2.70 × 10⁻³ mol; 2.7 × 10⁻³ mol
Common mistakes: Not converting cm³ to dm³; Incorrect multiplication
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Marking points
M1 · 1 mark · Converts 18.0 cm³ to dm³ and uses n = c × V (V in dm³) or shows appropriate working.
A1 · 1 mark · Correct answer: 0.00270 mol (or 2.70 × 10⁻³ mol, 2.7 × 10⁻³ mol).
Worked solution
First, convert the volume of HCl from cm³ to dm³:
Calculate moles of HCl: