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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.

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Presentation on theme: "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."— Presentation transcript:

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 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.

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 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 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%

3 6g of sodium hydrogen carbonate were dissolved in water and the volume made up to 250cm 3. 10.0cm 3 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. 28.30cm 3 of acid was needed. 2NaHCO 3 + H 2 SO 4  Na 2 SO 4 + 2CO 2 + 2H 2 O 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.

4 6g of sodium hydrogen carbonate were dissolved in water and the volume made up to 250cm 3. 10.0cm 3 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. 28.30cm 3 of acid was needed. 2NaHCO 3 + H 2 SO 4  Na 2 SO 4 + 2CO 2 + 2H 2 O Calculate: 1.The moles of sulfuric acid needed to neutralise the sodium hydrogen carbonate solution1.415 x 10 -3 2.The moles of sodium hydrogen carbonate in the original solution7.075 x 10 -2 3.The percentage purity of the sodium hydrogen carbonate. 99.1%

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

6 Washing soda crystals (10g) are dissolved and made up to 1dm 3 of solution. A 25.0cm 3 portion of the solution requires 35.80cm 3 of 0.05 moldm -3 sulphuric acid for neutralisation. Calculate the percentage by mass of sodium carbonate in the crystals. n H 2 SO 4 = 1.79 x 10 -3 n Na 2 CO 3 in original solution = 7.16 x 10 -2 % by mass Na 2 CO 3 = 75.9%


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