7.1 A standard solution is prepared by dissolving 10 g of sodium carbonate, Na2CO3(s), in 0.7 dm³ of water - English General - NSC Physical Sciences - Question 7 - 2024 - Paper 2

The pH of the sodium carbonate solution will be GREATER THAN 7.

This is because sodium carbonate is a salt derived from a weak acid (carbonic acid) and a strong base (sodium hydroxide), leading to basic conditions in the solution.

The dissociation of sodium carbonate in water can be represented as follows:

$Na_2CO_3(aq) + H_2O(l) \rightleftharpoons 2Na^+(aq) + HCO_3^-(aq) + OH^-(aq)$

Here, the generation of hydroxide ions ($OH^-$) confirms that the solution is basic, leading to a pH greater than 7.

The most suitable indicator for the titration of sodium carbonate with hydrochloric acid is indicator P, which has a pH range of 3.4 to 4.5.

This choice is appropriate because the equivalence point of the titration will be below pH 7 due to the weak base (sodium carbonate) reacting with a strong acid (hydrochloric acid).

A dilute acid is an acid that has a relatively low concentration of hydrogen ions ($H^+$) compared to a concentrated acid. This means that the acid is mixed with a significant amount of water, resulting in a solution where the pH is generally higher than that of a concentrated acid.

First, we calculate the moles of $[H^+]$ ions:

• Since H2SO4 is a diprotic acid, it produces 2 moles of $H^+$ for every mole:
• Moles of $H^+$ from H2SO4: $0.01 ext{ mol} \times 2 = 0.02 ext{ mol}$.
• Moles of $OH^-$ from KOH: 0.024 mol.

Since $OH^-$ exceeds $H^+$, we calculate the excess moles:

• Excess $OH^-$: $0.024 - 0.02 = 0.004 ext{ mol}$ in 0.2 dm³.

Concentration of $OH^-$: $[OH^-] = \frac{0.004}{0.2} = 0.02 ext{ mol/dm}^3$

Using the relation: $pOH = -\log[OH^-] = -\log(0.02) \approx 1.7$

Then, we find the pH using: $pH + pOH = 14 \Rightarrow pH = 14 - 1.7 = 12.3$