Acids, alkalis and titrations

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Question 10

A student performed a titration to find the concentration of a solution of calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ , using hydrochloric acid, $HCl\text{HCl}$ , of known concentration.

The student repeated the titration of calcium hydroxide solution with the hydrochloric acid and recorded these results:

Run	1	2	3	4
Burette reading after adding acid / $cm3\text{cm}^3$	26.50	25.80	25.35	26.25
Burette reading before adding acid / $cm3\text{cm}^3$	1.30	1.70	0.90	1.70
Volume of acid added / $cm3\text{cm}^3$	25.20	24.10	24.45	24.55

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 calcium hydroxide solution. The equation for the reaction is:

\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}

These are the student's experimental results:

Volume of calcium hydroxide solution: $cm325.0\text{ cm}^3$
Volume of hydrochloric acid: $cm322.40\text{ cm}^3$
Concentration of hydrochloric acid: $mol/dm30.100\text{ mol/dm}^3$

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

Step 1: $mol\text{amount of HCl} = \frac{0.100 \times 22.40}{100} = 0.0224\text{ mol}$
Step 2: $mol\text{amount of Ca(OH)}_2 = 0.0224 \times 2 = 0.0448\text{ mol}$
Step 3: $mol/dm3\text{concentration of Ca(OH)}_2 = \frac{0.0448}{22.40} \times 1000 = 2.00\text{ mol/dm}^3$

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 10

A student performed a titration to find the concentration of a solution of calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ , using hydrochloric acid, $HCl\text{HCl}$ , of known concentration.

The student repeated the titration of calcium hydroxide solution with the hydrochloric acid and recorded these results:

Run	1	2	3	4
Burette reading after adding acid / $cm3\text{cm}^3$	26.50	25.80	25.35	26.25
Burette reading before adding acid / $cm3\text{cm}^3$	1.30	1.70	0.90	1.70
Volume of acid added / $cm3\text{cm}^3$	25.20	24.10	24.45	24.55

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 calcium hydroxide solution. The equation for the reaction is:

\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}

These are the student's experimental results:

Volume of calcium hydroxide solution: $cm325.0\text{ cm}^3$
Volume of hydrochloric acid: $cm322.40\text{ cm}^3$
Concentration of hydrochloric acid: $mol/dm30.100\text{ mol/dm}^3$

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

Step 1: $mol\text{amount of HCl} = \frac{0.100 \times 22.40}{100} = 0.0224\text{ mol}$
Step 2: $mol\text{amount of Ca(OH)}_2 = 0.0224 \times 2 = 0.0448\text{ mol}$
Step 3: $mol/dm3\text{concentration of Ca(OH)}_2 = \frac{0.0448}{22.40} \times 1000 = 2.00\text{ mol/dm}^3$

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 10

Acids, alkalis and titrations Questions

Acids, alkalis and titrations

Question 10

Acids, alkalis and titrations Questions