8462/1H June 2021

9 ​​ This question is about acids.

Hydrogen chloride and ethanoic acid both dissolve in water. All hydrogen chloride molecules ionise in water.

Approximately​​ 1%​​ of​​ ethanoic​​ acid​​ molecules​​ ionise​​ in​​ water.

9-1  ​​​​ A solution is made by dissolving 1 g of hydrogen chloride in 1 dm3 of water.

Which​​ is​​ the​​ correct​​ description​​ of​​ this​​ solution?[1​​ mark]

 

ANSWER

 

a dilute solution of a strong acid

 

 

 

 

9-2  ​​​​ Which​​ solution​​ would​​ have​​ the​​ lowest​​ pH?  ​​ ​​ ​​ ​​ ​​ ​​​​ [1​​ mark]

 

ANSWER

1.0 mol/dm3​​ hydrogen chloride solution.

A​​ student​​ investigated​​ the​​ concentration​​ of​​ a​​ solution​​ of​​ sodium​​ hydroxide​​ by​​ titration with a 0.0480 mol/dm3​​ ethanedioic acid solution.

This​​ is​​ the​​ method​​ used.

  • Measure​​ 25.0​​ cm3​​ of​​ the​​ sodium​​ hydroxide​​ solution​​ into​​ a​​ conical​​ flask​​ using​​ a​​ 25.0​​ cm3​​ pipette.

  • Add​​ two​​ drops​​ of​​ indicator​​ to​​ the​​ sodium​​ hydroxide​​ solution.

  • Fill​​ a​​ burette​​ with​​ the​​ 0.0480​​ mol/dm3​​ ethanedioic​​ acid​​ solution​​ to​​ the​​ 0.00​​ cm3​​ mark.

  • Add​​ the​​ ethanedioic​​ acid​​ solution​​ to​​ the​​ sodium​​ hydroxide​​ solution​​ until​​ the indicator changes colour.

  • Read​​ the​​ burette​​ to​​ find​​ the​​ volume​​ of​​ the​​ ethanedioic​​ acid​​ solution​​ used.

9-3  ​​​​ Suggest two improvements to the method that would increase the accuracy​​ of​​ the​​ result.​​   ​​ ​​ ​​ ​​​​ [2​​ marks]

ANSWER

  • swirl the solution,

  • Use of white tile (under the flask) which will indicator clear visibility of color change;

  • Add ethanedioic acid drop wise near the endpoint;​​ 

  • Repeat and calculate mean.

 

9-4  ​​​​ Ethanedioic acid is a solid at room​​ temperature.

Calculate the mass of ethanedioic acid (H2C2O4) needed to make 250 cm3 of a solution with concentration 0.0480 mol/dm3

Relative​​ formula​​ mass​​ (Mr): H2C2O4​​ =​​ 90  ​​ ​​ ​​ ​​ ​​​​ [2​​ marks]

 

ANSWER

Since the concentration is given in mol/dm3,​​ therefore,​​ 

Conc. gmdm3=Mr×conc.moldm3

=90×0.0480

 

Now by using,​​ 

mass in gm=conc. gmdm3×volume(dm3)

=4.32×2501000

mass in gm ofethanedioic acid (H2C2O4)=1.08 g

 

The student found that 25.0 cm3 of the sodium hydroxide solution was neutralised by 15.00 cm3 of the 0.0480 mol/dm3 ethanedioic acid solution.

The equation for the reaction is:

H2C2O4​​ +​​ 2​​ NaOH​​ ​​ Na2C2O4​​ +​​ 2​​ H2O

9-5  ​​​​ Calculate​​ the​​ concentration​​ of​​ the​​ sodium​​ hydroxide​​ solution​​ in​​ mol/dm3.  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ [3​​ marks]

ANSWER

Step1: Calculating Moles of ethanedioic​​ acid​​ solution,

moles=conc.moldm3×volume(dm3)

moles ofethanedioic acid solution. =0.0480×151000

 

moles ofethanedioic acid solution.=0.00072mol

Step 2:​​ Now seeing the balanced equation, we need to identify​​ molar ​​ ratio

 

H2C2O4​​ : ​​ NaOH = 1:2

Therefore, moles of NaOH is​​ =2×0.00072mol

 

moles of NaOH is =0.00144mol

 

=0.00144251000