The table for burette readings is empty. To complete the calculations, I will assume an average titre value for the acid. Assumption: Let the average titre value (volume of acid used) be $\text{22.50 cm}^3$. Given Information: Solution B: Contains 0.02 mol of sodium hydroxide (NaOH). Volume of solution B (stock) = 250 cm³. Solution A: Tetraoxosulphate (IV) acid ($H_2SO_4$). Volume of solution B used in titration ($V_b$) = 25 cm³. Average titre value (volume of acid A used, $V_a$) = 22.50 cm³ (assumed). Step 1: Calculate the concentration of the base (Solution B). Moles of NaOH = 0.02 mol Volume of stock solution B = 250 cm³ = 0.250 dm³ $$M_b = \frac{\text{Moles of NaOH}}{\text{Volume of solution B (dm}^3)} = \frac{0.02 \text{ mol}}{0.250 \text{ dm}^3} = 0.08 \text{ mol/dm}^3$$ So, the molar concentration of the base ($M_b$) is $0.08 \text{ mol/dm}^3$. Step 2: Write the balanced chemical equation for the reaction. The reaction is between sulfuric acid ($H_2SO_4$) and sodium hydroxide (NaOH). $$\text{H}_2\text{SO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{H}_2\text{O} (l)$$ From the balanced equation: Moles of acid ($n_a$) = 1 Moles of base ($n_b$) = 2 --- i) The molar concentration of the acid. Step 1: Convert volumes to dm³. $V_a = 22.50 \text{ cm}^3 = 0.02250 \text{ dm}^3$ $V_b = 25 \text{ cm}^3 = 0.025 \text{ dm}^3$ Step 2: Use the titration formula. $$\frac{M_aV_a}{n_a} = \frac{M_bV_b}{n_b}$$ Rearrange to solve for $M_a$: $$M_a = \frac{M_bV_b n_a}{V_a n_b}$$ Substitute the known values: $$M_a = \frac{(0.08 \text{ mol/dm}^3) \times (0.025 \text{ dm}^3) \times 1}{(0.02250 \text{ dm}^3) \times 2}$$ $$M_a = \frac{0.002}{0.045}$$ $$M_a = 0.04444... \text{ mol/dm}^3$$ Rounding to three significant figures: $$M_a = \boxed{\text{0.0444 mol/dm}^3}$$ ii) The concentration mass of the acid. Step 1: Calculate the molar mass of $H_2SO_4$. Atomic masses: H = 1.01 g/mol, S = 32.07 g/mol, O = 16.00 g/mol $$MM_{H_2SO_4} = (2 \times 1.01) + 32.07 + (4 \times 16.00)$$ $$MM_{H_2SO_4} = 2.02 + 32.07 + 64.00 = 98.09 \text{ g/mol}$$ Step 2: Calculate the mass concentration. $$\text{Mass Concentration} = M_a \times MM_{H_2SO_4}$$ $$\text{Mass Concentration} = 0.04444 \text{ mol/dm}^3 \times 98.09 \text{ g/mol}$$ $$\text{Mass Concentration} = 4.359 \text{ g/dm}^3$$ Rounding to three significant figures: $$\text{Mass Concentration} = \boxed{\text{4.36 g/dm}^3}$$ iii) From a balanced chemical equation, determine the mole ratio of the base. Step 1: Write the balanced chemical equation. $$\text{H}_2\text{SO}_4 (aq) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2\text{H}_2\text{O} (l)$$ Step 2: Determine the mole ratio. The stoichiometric coefficient for the base (NaOH) is 2. The mole ratio of acid to base ($H_2SO_4 : NaOH$) is $1:2$. The mole ratio of the base (NaOH) in the reaction is $\boxed{2}$. iv) Sketch the titration curve of the acid-base titration against the pH. This is a titration of a strong base (NaOH) with a strong acid ($H_2SO_4$). The base is initially in the conical flask, and the acid is added from the burette. Characteristics of the curve: Initial pH: High (basic), as the solution starts with NaOH. Initial pH of 0.08 M NaOH: $pOH = -\log(0.08) = 1.10$, so $pH = 14 - 1.10 = 12.90$. Equivalence point: At pH 7.0, because it's a strong acid-strong base titration. Shape: The pH decreases gradually, then sharply drops around the equivalence point (pH 7), and then continues to decrease gradually as excess acid is added. Final pH: Low (acidic), as excess $H_2SO_4$ is added. Here is a sketch of the titration curve: $$ \begin{tikzpicture}[scale=0.8] \draw[->] (0,0) -- (10,0) node[below] {Volume of Acid (cm$^3$)}; \draw[->] (0,0) -- (0,10) node