8462/1H June 2018

This​​ question​​ is​​ about​​ acids​​ and​​ alkalis.​​ 

Dilute hydrochloric acid is a strong​​ acid.

Explain why an acid can be described as both strong and dilute. 2 marks]

ANSWER

A concentrated acid is one in which many acid molecules are dissolved in a set volume of solution, while a dilute acid will have very few molecules per unit volume.​​ 

A 1.0 x 10–3​​ mol/dm3​​ solution of hydrochloric acid has a pH of 3.0.​​ 

What is the pH of a 1.0 x 10–5​​ mol/dm3​​ solution of hydrochloric acid?​​ 

[1 mark]

ANSWER

pH will be 5

A​​ student​​ titrated​​ 25.0​​ cm3​​ portions​​ of​​ dilute​​ sulfuric​​ acid​​ with​​ a​​ 0.105​​ mol/dm3​​ sodium​​ hydroxide​​ solution.

 

Table​​ 4​​ shows​​ the​​ student’s​​ results.

Table​​ 4

The equation for the reaction is:

Calculate the concentration of the sulfuric acid in mol/dm3.​​ 

Use only the student’s concordant results.​​ 

Concordant results are those within 0.10 cm3​​ of each other.  ​​ ​​ ​​​​ 

[5 marks]

ANSWER

Keeping in view the concordant results, so we choose values of titrations 3, 4 & 5;

Average of Volume of NaOH by using only titration values 3,4 &5=22.10+22.15+22.153

=22.13cm3

Moles of NaOH=22.131000×0.105

=0.002324mol

By seeing the given equation, the reacting reaction of NaOH: H2SO4​​ = 2:1

Therefore,​​ 

Moles of H2SO4=12×0.002324

=0.001162mol

=0.001162mol251000

 

 

Explain why the student should use a pipette to measure the dilute sulfuric acid and a burette to measure the sodium hydroxide solution.   ​​ 

[2 marks]

ANSWER

Pipette measures a fixed volume accurately but burette measures variable volume.

Calculate the mass of sodium hydroxide in 30.0 cm3​​ of a 0.105 mol/dm3​​ solution.

Relative formula mass (Mr): NaOH = 40     

[2 marks]

ANSWER

moles=0.105×(301000)

=0.00315 mol

mass in gm=mol×Mr

=0.00315×40

=0.126gm