Acids, alkalis and titrations

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Question 24

A student performed a titration to find the concentration of a solution of strontium hydroxide, $Sr(OH)2\text{Sr(OH)}_2$ , using nitric acid, $HNO3\text{HNO}_3$ , of known concentration.

The student repeated the titration of strontium hydroxide solution with the nitric acid and recorded these results:

Run	1	2	3	4
Burette reading after adding acid / $cm3\text{cm}^3$	24.80	25.55	26.10	25.95
Burette reading before adding acid / $cm3\text{cm}^3$	1.30	1.15	0.80	1.45
Volume of acid added / $cm3\text{cm}^3$	23.50	24.40	25.30	24.50

The average (mean) volume of acid should be calculated using only concordant results. Concordant results are those volumes that differ from each other by $cm30.20\text{ cm}^3$ or less.

Identify the concordant results by stating which runs should be selected.

[1]

Calculate the average (mean) volume of acid added using your selected results.

[1]

The student used the same method to find the concentration of another strontium hydroxide solution. The equation for the reaction is:

\text{Sr(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Sr(NO}_3)_2 + 2\text{H}_2\text{O}

These are the student's experimental results:

Volume of strontium hydroxide solution: $cm320.0\text{ cm}^3$
Volume of nitric acid: $cm321.60\text{ cm}^3$
Concentration of nitric acid: $mol/dm30.125\text{ mol/dm}^3$

The student used these results to calculate the concentration of the strontium hydroxide solution:

Step 1: $mol\text{amount of HNO}_3 = \frac{0.125 \times 21.60}{100} = 0.0270\text{ mol}$
Step 2: $mol\text{amount of Sr(OH)}_2 = 0.0270 \times 2 = 0.0540\text{ mol}$
Step 3: $mol/dm3\text{concentration of Sr(OH)}_2 = \frac{0.0540}{21.60} \times 1000 = 2.50\text{ mol/dm}^3$

There is one mistake in each step of the calculation. What correction should the student make in:

Step 1?

[1]

Step 2?

[1]

Step 3?

[1]

Acids, alkalis and titrations

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Question 24

A student performed a titration to find the concentration of a solution of strontium hydroxide, $Sr(OH)2\text{Sr(OH)}_2$ , using nitric acid, $HNO3\text{HNO}_3$ , of known concentration.

The student repeated the titration of strontium hydroxide solution with the nitric acid and recorded these results:

Run	1	2	3	4
Burette reading after adding acid / $cm3\text{cm}^3$	24.80	25.55	26.10	25.95
Burette reading before adding acid / $cm3\text{cm}^3$	1.30	1.15	0.80	1.45
Volume of acid added / $cm3\text{cm}^3$	23.50	24.40	25.30	24.50

The average (mean) volume of acid should be calculated using only concordant results. Concordant results are those volumes that differ from each other by $cm30.20\text{ cm}^3$ or less.

Identify the concordant results by stating which runs should be selected.

[1]

Calculate the average (mean) volume of acid added using your selected results.

[1]

The student used the same method to find the concentration of another strontium hydroxide solution. The equation for the reaction is:

\text{Sr(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Sr(NO}_3)_2 + 2\text{H}_2\text{O}

These are the student's experimental results:

Volume of strontium hydroxide solution: $cm320.0\text{ cm}^3$
Volume of nitric acid: $cm321.60\text{ cm}^3$
Concentration of nitric acid: $mol/dm30.125\text{ mol/dm}^3$

The student used these results to calculate the concentration of the strontium hydroxide solution:

Step 1: $mol\text{amount of HNO}_3 = \frac{0.125 \times 21.60}{100} = 0.0270\text{ mol}$
Step 2: $mol\text{amount of Sr(OH)}_2 = 0.0270 \times 2 = 0.0540\text{ mol}$
Step 3: $mol/dm3\text{concentration of Sr(OH)}_2 = \frac{0.0540}{21.60} \times 1000 = 2.50\text{ mol/dm}^3$

There is one mistake in each step of the calculation. What correction should the student make in:

Step 1?

[1]

Step 2?

[1]

Step 3?

[1]

Acids, alkalis and titrations

Question 24

Acids, alkalis and titrations Questions

Acids, alkalis and titrations

Question 24

Acids, alkalis and titrations Questions