I will solve the questions based on the provided image, interpreting the blurry text as accurately as possible. I have identified two new titration problems.

$$\text{Moles of Na}_2\text{CO}_3 = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of Na}_2\text{CO}_3 = 0.02 \text{ mol/dm}^3 \times 0.020 \text{ dm}^3 = 0.0004 \text{ mol}$$

The number of moles of sodium carbonate used is $\boxed{0.0004 \text{ mol}}$.

g) (ii) Determine the number of moles of hydrochloric acid used.
From the balanced equation, 1 mole of $\text{Na}_2\text{CO}_3$ reacts with 2 moles of $\text{HCl}$.

$$\text{Moles of HCl} = 2 \times \text{Moles of Na}_2\text{CO}_3$$ $$\text{Moles of HCl} = 2 \times 0.0004 \text{ mol} = 0.0008 \text{ mol}$$

The number of moles of hydrochloric acid used is $\boxed{0.0008 \text{ mol}}$.

g) (iii) Determine the concentration of the acid in moles per litre.

$$\text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}}$$ $$\text{Concentration of HCl} = \frac{0.0008 \text{ mol}}{0.020 \text{ dm}^3} = 0.04 \text{ mol/dm}^3$$

The concentration of the acid in moles per litre is $\boxed{0.04 \text{ mol/dm}^3}$.

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Question 3 (Titration 2):

Given:

• Concentration of $\text{NaOH}$ solution = 12.4 g/dm$^3$
• Volume of $\text{NaOH}$ solution used = 25 cm$^3$
• Volume of sulphuric (VI) acid ($\text{H}_2\text{SO}_4$) solution used = 15 cm$^3$
• Molar masses: Na=23.0, O=16.0, H=1.0, S=32.0

Reaction Equation:

$$\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$$

a) (i) Find the molarity of NaOH solution.