6
A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3 . The student then takes 25.0cm 3 of this solution and titrates it with dilute hydrochloric acid. He finds that he needs 25.80cm 3 of 0.200moldm -3 hydrochloric acid to fully neutralise the potassium hydroxide. KOH + HCl KCl + H 2 O Calculate: 1.The moles of HCl used to neutralise the KOH. 2.The moles of KOH in the original solution 3.The percentage purity of the potassium hydroxide.

A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

Embed Size (px)

Citation preview

Page 1: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm3.

The student then takes 25.0cm3 of this solution and titrates it with dilute hydrochloric acid. He finds that he needs 25.80cm3 of 0.200moldm-3 hydrochloric acid to fully neutralise the potassium hydroxide.

KOH + HCl KCl + H2O

Calculate:

1.The moles of HCl used to neutralise the KOH.

2.The moles of KOH in the original solution

3.The percentage purity of the potassium hydroxide.

Page 2: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm3.

The student then takes 25.0cm3 of this solution and titrates it with dilute hydrochloric acid. He finds that he needs 25.80cm3 of 0.200moldm-3 hydrochloric acid to fully neutralise the potassium hydroxide.

KOH + HCl KCl + H2O

Calculate:

1.The moles of HCl used to neutralise the KOH 5.16 x 10-3

2.The moles of KOH in the original solution 5.16 x 10-2

3.The percentage purity of the potassium hydroxide 96.5%

Page 3: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

6g of sodium hydrogen carbonate were dissolved in water and the volume made up to 250cm3.

10.0cm3 of this solution was pipetted into a conical flask, and sulphuric acid of concentration 0.0500moldm-3 was run in from the burette until the solution was neutralised. 28.30cm3 of acid was needed.

2NaHCO3 + H2SO4 Na2SO4 + 2CO2 + 2H2O

Calculate:

1.The moles of sulfuric acid needed to neutralise the sodium hydrogen carbonate solution

2.The moles of sodium hydrogen carbonate in the original solution

3.The percentage purity of the sodium hydrogen carbonate.

Page 4: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

6g of sodium hydrogen carbonate were dissolved in water and the volume made up to 250cm3.

10.0cm3 of this solution was pipetted into a conical flask, and sulphuric acid of concentration 0.0500moldm-3 was run in from the burette until the solution was neutralised. 28.30cm3 of acid was needed.

2NaHCO3 + H2SO4 Na2SO4 + 2CO2 + 2H2O

Calculate:

1.The moles of sulfuric acid needed to neutralise the sodium hydrogen carbonate solution 1.415 x 10-3

2.The moles of sodium hydrogen carbonate in the original solution 7.075 x 10-2

3.The percentage purity of the sodium hydrogen carbonate. 99.1%

Page 5: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

Washing soda crystals (10g) are dissolved and made up to 1dm3 of solution. A 25.0cm3 portion of the solution requires 35.80cm3 of 0.05 moldm-3 sulphuric acid for neutralisation. Calculate the percentage of sodium carbonate in the crystals.

Page 6: A student dissolves 3g of impure potassium hydroxide in water and makes the solution up to 250cm 3. The student then takes 25.0cm 3 of this solution and

Washing soda crystals (10g) are dissolved and made up to 1dm3 of solution. A 25.0cm3 portion of the solution requires 35.80cm3 of 0.05 moldm-3 sulphuric acid for neutralisation. Calculate the percentage by mass of sodium carbonate in the crystals.

n H2SO4 = 1.79 x 10-3

n Na2CO3 in original solution = 7.16 x 10-2

% by mass Na2CO3 = 75.9%