Home / iGCSE Chemistry 0620 Extended – 12.2 Acid–base titrations- Exam Style Questions Paper 4

iGCSE Chemistry 0620 Extended - 12.2 Acid–base titrations- Exam Style Questions Paper 4- New Syllabus

Question

A student makes crystals of the salt sodium sulfate, $Na_{2}SO_{4}$. The student reacts $0.200\text{ mol/dm}^{3}$ dilute sulfuric acid, $H_{2}SO_{4}(aq)$, with aqueous sodium hydroxide, $NaOH(aq)$.

The student uses the following steps.

  • step 1: The student places $40.0\text{ cm}^{3}$ of $NaOH(aq)$ into a conical flask. This volume contains $0.0100\text{ moles}$ of $NaOH$.
  • step 2: The student adds a few drops of methyl orange indicator to the $NaOH(aq)$ in the conical flask.
  • step 3: The student adds $0.200\text{ mol/dm}^{3}\;H_{2}SO_{4}(aq)$ to the flask until the end-point is reached.
  • step 4: The student transfers the mixture from the conical flask to an evaporating basin and obtains dry crystals.
(a) Complete the symbol equation for the reaction. Include state symbols.
$H_{2}SO_{4}(aq) + 2NaOH(aq) \rightarrow Na_{2}SO_{4}(\dots) + \dots\dots\dots(\dots)$
(b) State the type of exothermic reaction taking place.
(c) Calculate the concentration of $NaOH(aq)$ used in step 1.
(d) Name the item of apparatus the student uses to add $H_{2}SO_{4}(aq)$ in step 3.
(e) Calculate the volume of $H_{2}SO_{4}(aq)$, in $\text{cm}^{3}$, added in step 3.
(f) State the colour change observed in step 3 (from / to).
(g) The dry crystals formed in step 4 are coloured and not white. This is because the student should do an additional step between step 3 and step 4. Suggest what the student should do in this additional step to produce white crystals.
(h) In step 4, the student gently heats the solution in the evaporating basin until the solution is saturated. The student then stops heating and leaves the hot solution to cool. Crystals start to appear.
(i) Explain the term saturated solution.
(ii) Explain why crystals start to appear as the hot solution cools.
(iii) Suggest the effect, if any, on the mass of crystals collected in step 4 if the solution in the evaporating basin is allowed to dry without gentle heating.

Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):

• Topic $7.1$ — The characteristic properties of acids and bases (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $12.2$ — Acid-base titrations (Parts $\mathrm{(d)}$, $\mathrm{(f)}$, $\mathrm{(g)}$)
• Topic $3.3$ — The mole and the Avogadro constant (Parts $\mathrm{(c)}$, $\mathrm{(e)}$)
• Topic $12.4$ — Separation and purification (Part $\mathrm{(h)}$)

▶️ Answer/Explanation

(a) $H_{2}SO_{4}(aq) + 2NaOH(aq) \rightarrow Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

(b) Neutralisation

(c) $0.250\text{ mol/dm}^{3}$
Calculation: Concentration = $\frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.0100}{40.0 \div 1000} = 0.250\text{ mol/dm}^{3}$

(d) Burette

(e) Volume = $25\text{ cm}^{3}$
Calculation: From the balanced equation, $1\text{ mol}$ of $H_{2}SO_{4}$ reacts with $2\text{ mol}$ of $NaOH$.
Moles of $H_{2}SO_{4}$ needed = $\frac{0.0100}{2} = 0.00500\text{ mol}$.
Volume = $\frac{\text{moles}}{\text{concentration}} = \frac{0.00500}{0.200} = 0.025\text{ dm}^{3} = 25\text{ cm}^{3}$.

(f) From yellow to orange.
Explanation: Methyl orange is yellow in alkaline solutions ($NaOH$) and turns orange at the neutral end-point.

(g) Repeat steps 1 and 3 exactly, but without adding the indicator.
Explanation: The indicator remains in the solution and stains the crystals. A titration must be repeated without the indicator using the exact known volumes to get pure white crystals.

(h)(i) A solution containing the maximum concentration of a solute dissolved in the solvent at a specified temperature.
(h)(ii) The solubility of the solid decreases as the temperature decreases.
(h)(iii) None.

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