A student is performing a titration. Solution D is 0.100 mol/dm³ NaOH and Solution E is dilute HCl of unknown concentration. 25.0 cm³ of Solution D is titrated against Solution E using methyl orange i

Part (a): This part describes the experimental setup for the titration. Solution D ($0.100 \text{ mol/dm}^3 \text{ NaOH}$) is the standard solution, and Solution E (dilute $\text{HCl}$) is the analyte of unknown concentration. Solution E is placed in the burette and titrated against $25.0 \text{ cm}^3$ portions of Solution D using methyl orange indicator.

Part (b):
To tabulate results and calculate the average volume of acid used, experimental data is required. Since no titration data is provided in the image, we will assume an average titre volume for the acid (Solution E) to demonstrate the subsequent calculations.

• Assumption: Let the average volume of acid (Solution E) used be $20.00 \text{ cm}^3$.

Part (c):

Step 1: Write the balanced chemical equation for the reaction between sodium hydroxide ($\text{NaOH}$) and hydrochloric acid ($\text{HCl}$).

$$\text{NaOH} (aq) + \text{HCl} (aq) \to \text{NaCl} (aq) + \text{H}_2\text{O} (l)$$

Part (d):

Step 1: Identify the known values and the assumed average titre.
Concentration of base, $C_B = C_D = 0.100 \text{ mol/dm}^3$
Volume of base, $V_B = V_D = 25.0 \text{ cm}^3$
Volume of acid, $V_A = V_E = 20.00 \text{ cm}^3$ (assumed from part b)
From the balanced equation, the mole ratio of acid to base is $n_A : n_B = 1:1$.

Step 2: Use the titration formula to calculate the concentration of the acid ($C_A$).

$$\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$$ $$C_A = \frac{C_B V_B n_A}{V_A n_B}$$ $$C_A = \frac{(0.100 \text{ mol/dm}^3) \times (25.0 \text{ cm}^3) \times 1}{(20.00 \text{ cm}^3) \times 1}$$ $$C_A = \frac{2.50}{20.00} \text{ mol/dm}^3$$ $$C_A = 0.125 \text{ mol/dm}^3$$

The concentration of E in $\text{mol/dm}^3$ is $\boxed{\text{0.125 mol/dm}^3}$.

Step 3: Calculate the molar mass of $\text{HCl}$.
Molar mass of $\text{H} = 1.008 \text{ g/mol}$
Molar mass of $\text{Cl} = 35.453 \text{ g/mol}$
Molar mass of $\text{HCl} = 1.008 + 35.453 = 36.461 \text{ g/mol}$

Step 4: Convert the concentration from $\text{mol/dm}^3$ to $\text{g/dm}^3$.
Concentration in $\text{g/dm}^3 = \text{Concentration in mol/dm}^3 \times \text{Molar mass}$
Concentration in $\text{g/dm}^3 = 0.125 \text{ mol/dm}^3 \times 36.461 \text{ g/mol}$
Concentration in \text{g/dm}^3 = 4.557625 \text{ g/dm}^3$$ The concentration of E in \text{g/dm}^3$is$\boxed{\text{4.56 g/dm}^3}$ (to 3 significant figures).

Part (e):

Step 1: Determine the mass of pure $\text{HCl}$ in $1.0 \text{ dm}^3$ of solution E.
From part (d), the concentration of pure $\text{HCl}$ is $4.557625 \text{ g/dm}^3$.
So, in $1.0 \text{ dm}^3$ of solution, the mass of pure $\text{HCl}$ is $4.557625 \text{ g}$.

Step 2: Calculate the percentage purity.
Given that the impure acid used to prepare the solution was $4.0 \text{ g}$ of $\text{HCl}$ in $1.0 \text{ dm}^3$.

$$\text{Percentage purity} = \frac{\text{Mass of pure HCl}}{\text{Mass of impure HCl}} \times 100\%$$ $$\text{Percentage purity} = \frac{4.557625 \text{ g}}{4.0 \text{ g}} \times 100\%$$ $$\text{Percentage purity} = 113.940625\%$$

This result (over 100%) indicates that the assumed average titre volume or the given mass of impure HCl might be inconsistent with a real-world scenario where purity cannot exceed 100%. However, based on the numbers provided and calculated, this is the mathematical result.
The percentage purity of the acid is $\boxed{\text{113.9\%}}$ (to 1 decimal place).

Part (f):

• (i) Dilution factor

Step 1: Identify the initial and final volumes.
Initial volume, $V_1 = 100 \text{ cm}^3$
Final volume, $V_2 = 500 \text{ cm}^3$

Step 2: Calculate the dilution factor.

$$\text{Dilution factor} = \frac{\text{Final volume}}{\text{Initial volume}}$$ $$\text{Dilution factor} = \frac{500 \text{ cm}^3}{100 \text{ cm}^3}$$ $$\text{Dilution factor} = 5$$

The dilution factor is $\boxed{5}$.

• (ii) Concentration of the new solution in $\text{mol/dm}^3$

Step 1: Use the dilution formula $C_1 V_1 = C_2 V_2$.
Initial concentration, $C_1 = 0.125 \text{ mol/dm}^3$ (from part d)
Initial volume, $V_1 = 100 \text{ cm}^3$
Final volume, $V_2 = 500 \text{ cm}^3$
Final concentration, $C_2 = ?$

Step 2: Calculate the new concentration ($C_2$).

$$C_2 = \frac{C_1 V_1}{V_2}$$ $$C_2 = \frac{(0.125 \text{ mol/dm}^3) \times (100 \text{ cm}^3)}{500 \text{ cm}^3}$$ $$C_2 = \frac{12.5}{500} \text{ mol/dm}^3$$ $$C_2 = 0.025 \text{ mol/dm}^3$$

The concentration of the new solution is $\boxed{\text{0.025 mol/dm}^3}$.