Question 1 Report
This diagram is used in a school laboratory to show a pH probe connected to a data logger. A student adds sodium hydroxide solution in small portions to 25.0 cm3 of sulfuric acid. The data logger records the pH after each addition, rather than relying on an indicator colour change.
| total volume of sodium hydroxide solution added / cm3 | pH |
|---|---|
| 0.0 | 1.2 |
| 12.0 | 1.8 |
| 24.0 | 3.4 |
| 25.0 | 7.0 |
| 26.0 | 10.5 |
(a) State the pH of a neutral solution. [1]
(b) Use Table 1 to identify the volume of sodium hydroxide solution at the neutralisation point. [1]
(c) Explain why the pH changes very rapidly when the volume added is close to 25.0 cm3. [2]
(d) Calculate the concentration of the sulfuric acid. The sodium hydroxide solution has concentration 0.200 mol/dm3. Use the equation H2SO4 + 2NaOH → Na2SO4 + 2H2O. [5]
(e) Give one advantage of using a pH probe rather than an indicator for a strongly coloured acid solution. [2]
(a) A neutral solution has pH 7. [1]
(b) The neutralisation point is at 25.0 cm3 of sodium hydroxide added, because the recorded pH is 7.0. [1]
(c) Close to 25.0 cm3, nearly all the hydrogen ions from the acid have reacted with hydroxide ions from the alkali. A very small further addition of sodium hydroxide then leaves hydroxide ions in excess, so the solution rapidly becomes alkaline and its pH rises sharply. [2]
(d) First find the amount of sodium hydroxide at the neutralisation point:
\[n(\mathrm{NaOH})=cV=0.200\times0.0250=0.00500\text{ mol}\]
The equation shows \(2\) mol NaOH react with \(1\) mol \(\mathrm{H_2SO_4}\):
\[n(\mathrm{H_2SO_4})=\frac{0.00500}{2}=0.00250\text{ mol}\]
The acid volume is \(25.0\text{ cm}^3=0.0250\text{ dm}^3\).
\[c(\mathrm{H_2SO_4})=\frac{n}{V}=\frac{0.00250}{0.0250}=0.100\text{ mol dm}^{-3}\]
The sulfuric acid concentration is \(0.100\text{ mol dm}^{-3}\). [5]
(e) A pH probe gives a numerical pH reading, so no judgement of an indicator colour is needed. This is particularly useful when the acid solution is strongly coloured. [2]
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