Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Section
The standard calculation
19 practice questions
Practice questions
1–10 of 19
Question 1
1 mark · Easy
Answer and see the solution →State the first step you should always take when calculating the concentration of a solution using the three-step titration calculation method.
Question 2
3 marks · Easy
Answer and see the solution →A student is titrating 25.0 cm³ of sodium hydroxide solution with hydrochloric acid. The concentration of sodium hydroxide is 0.200 mol/dm³.
(a)[2 marks]Calculate the number of moles of sodium hydroxide in 25.0 cm³.
(b)[1 mark]State the equation you would use for this calculation.
(c)[0 marks]Give the unit for the concentration used in the calculation.
Question 3
2 marks · Easy
Answer and see the solution →The equation for the reaction is:
In a titration, 35.0 cm³ of 0.300 mol/dm³ potassium hydroxide are used. Calculate the number of moles of sulfuric acid that will react with this amount, using the equation.
Question 4
2 marks · Easy
Answer and see the solution →Explain, using the three-step titration calculation method, how you can find the concentration of an unknown acid after titrating it with a solution of known concentration.
Question 5
3 marks · Easy
Answer and see the solution →A student has the following titration results:
Titration Volume of NaOH (cm³) Concentration of NaOH (mol/dm³) 1 22.0 0.150 2 21.5 0.150 3 (repeat) 21.7 0.150 The student used 25.0 cm³ of hydrochloric acid of unknown concentration each time. Use the results to calculate the mean number of moles of sodium hydroxide used to neutralise the acid.
Question 6
1 mark · Easy
Answer and see the solution →Write the balanced symbol equation for the reaction between potassium hydroxide and nitric acid.
Question 7
3 marks · Easy
Answer and see the solution →A student uses 28.0 cm³ of hydrochloric acid to neutralise 25.0 cm³ of sodium hydroxide solution with a concentration of 0.120 mol/dm³. The equation for the reaction is:
(a)[1 mark]Calculate the moles of sodium hydroxide used.
(b)[1 mark]State how many moles of hydrochloric acid react with this amount, using the equation.
(c)[1 mark]Calculate the concentration of hydrochloric acid in mol/dm³.
Question 8
2 marks · Easy
Answer and see the solution →A titration uses 20.0 cm³ of 0.175 mol/dm³ nitric acid ( [3m[33mHNO[0m[1;31m[0m[0;32m3[0m ) solution for neutralisation.
Calculate the number of moles of nitric acid used. Show your steps.
Question 9
2 marks · Normal
Answer and see the solution →A 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.
Question 10
4 marks · Normal
Answer and see the solution →Ethanoic acid ( ) reacts with sodium hydroxide ( ) according to the equation:
A student titrates 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide with ethanoic acid. On average, 21.0 cm³ of ethanoic acid is needed to neutralise the alkali.
(a)[1 mark]State the ratio in which sodium hydroxide reacts with ethanoic acid.
(b)[1 mark]Calculate the number of moles of sodium hydroxide present in 25.0 cm³.
(c)[1 mark]Calculate the number of moles of ethanoic acid used in the reaction.
(d)[1 mark]Calculate the concentration of the ethanoic acid.