SC14d Titrations and Calculations

Question 1

25cm3 of sodium hydroxide solution as titrated against 0.100mol dm3 hydrochloric acid. An average of 20cm3 of the acid neutralised the alkali. Calculate the concentration of the sodium hydroxide solution.
HCl + NaOH → NaCl + H2O

Answer 1

1. Calculate the number of moles of the solution of known volume and concentration
   moles of HCl = 0.100mol dm3 x 0.2dm3 = 0.002mol
2. Use the balance equation to work out the number of moles of alkali that reacted
   0.002 HCl : 0.002 NaOH
3. Calculate the concentration of the sodium hydroxide solution
   volume of NaOH = 25/1000 = 0.0250dm3
   concentration of NaOH = 0.002/0.0250 = 0.08mol dm3

Question 2

25.0cm3 of potassium hydroxide solutions reacted with 23.3cm3 of 0.100moldm3 hydrochloric acid. Calculate the concentration of this potassium hydroxide solution in mol dm3 (3 marks)
HCl + KOH → KCl + H2O

Answer 2

Moles ratio HCl : KOH = 1:1
moles HCl = 0.100 x 23.3/1000 = 0.00233
moles KOH = 0.00233
concentration KOH = 0.00233 x 1000/25.0 = 0.932mol dm3