Please this is chemistry assignment

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Question 4:

• Given:

  • Concentration of acid D ($C_A$) = $0.05 \text{moldm}^{-3}$ $\text{H}_2\text{SO}_4$
  • Volume of base E ($V_B$) = $25.0 \text{cm}^3$ impure $\text{KOH}$
  • Initial concentration of impure $\text{KOH}$ = $7.50 \text{gdm}^{-3}$
  • Balanced equation: $2\text{KOH} + \text{H}_2\text{SO}_4 \longrightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$
  • Mole ratio of $\text{KOH} : \text{H}_2\text{SO}_4$ is $n_B : n_A = 2:1$.

• Titration Readings:

  • 1st titration: $24.50 - 0.00 = 24.50 \text{cm}^3$
  • 2nd titration: $46.70 - 22.50 = 24.20 \text{cm}^3$
  • 3rd titration: $24.10 - 0.00 = 24.10 \text{cm}^3$

(i) Average volume of the acid used. Step 1: Identify concordant readings. The 2nd and 3rd titrations ($24.20 \text{cm}^3$ and $24.10 \text{cm}^3$) are concordant. Step 2: Calculate the average volume of acid ($V_A$). $V_A = \frac{24.20 \text{ cm}^3 + 24.10 \text{ cm}^3}{2}$ $V_A = \frac{48.30 \text{ cm}^3}{2}$ $V_A = \boxed{\text{24.15 cm}^3}$

(ii) Concentration of E in $\text{moldm}^{-3}$ Step 1: Use the titration formula $\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$ to find the concentration of the base ($C_B$). $C_B = \frac{n_B C_A V_A}{n_A V_B}$ Step 2: Substitute the known values. $C_B = \frac{2 \times 0.05 \text{ moldm}^{-3} \times 24.15 \text{ cm}^3}{1 \times 25.0 \text{ cm}^3}$ $C_B = \frac{2.415 \text{ moldm}^{-3}}{25.0}$ $C_B = \boxed{\text{0.0966 moldm}^{-3}}$

(iii) Concentration of E in $\text{gdm}^{-3}$ Step 1: Calculate the molar mass of $\text{KOH}$. Molar mass of $\text{KOH} = \text{K} + \text{O} + \text{H} = 39 + 16 + 1 = 56 \text{ g/mol}$. Step 2: Convert the concentration from $\text{moldm}^{-3}$ to $\text{gdm}^{-3}$. $\text{Concentration in gdm}^{-3} = \text{Concentration in moldm}^{-3} \times \text{Molar mass}$ $\text{Concentration of pure KOH} = 0.0966 \text{ moldm}^{-3} \times 56 \text{ g/mol}$ $\text{Concentration of pure KOH} = \boxed{\text{5.41 gdm}^{-3}}$

(iv) Percentage of impurity in an impure solution E Step 1: Calculate the mass of impurity. $\text{Mass of impurity} = \text{Concentration of impure KOH} - \text{Concentration of pure KOH}$ $\text{Mass of impurity} = 7.50 \text{ gdm}^{-3} - 5.41 \text{ gdm}^{-3}$ $\text{Mass of impurity} = 2.09 \text{ gdm}^{-3}$ Step 2: Calculate the percentage of impurity. $\text{Percentage impurity} = \frac{\text{Mass of impurity}}{\text{Concentration of impure KOH}} \times 100\%$ $\text{Percentage impurity} = \frac{2.09 \text{ gdm}^{-3}}{7.50 \text{ gdm}^{-3}} \times 100\%$ $\text{Percentage impurity} = 0.27866 \times 100\%$ \text{Percentage impurity} = \boxed{\text{27.9%}}

(v) Give reason why an indicator has to be added to either an acid solution or a base solution during titration An indicator is added to a titration mixture to visually signal the equivalence point of the reaction. It does this by changing color when the $\text{pH}$ of the solution reaches a specific range, which corresponds to the point where the acid and base have completely neutralized each other. Without an indicator, it would be impossible to observe this point.