This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Hey zionenergy4, good to see you again. Here are the solutions to your chemistry assignment: Question 4: Given: Concentration of acid D (C_A) = 0.05 moldm^-3 H_2SO_4 Volume of base E (V_B) = 25.0 cm^3 impure KOH Initial concentration of impure KOH = 7.50 gdm^-3 Balanced equation: 2KOH + H_2SO_4 K_2SO_4 + 2H_2O Mole ratio of KOH : H_2SO_4 is n_B : n_A = 2:1. Titration Readings: 1st titration: 24.50 - 0.00 = 24.50 cm^3 2nd titration: 46.70 - 22.50 = 24.20 cm^3 3rd titration: 24.10 - 0.00 = 24.10 cm^3 (i) Average volume of the acid used. Step 1: Identify concordant readings. The 2nd and 3rd titrations (24.20 cm^3 and 24.10 cm^3) are concordant. Step 2: Calculate the average volume of acid (V_A). V_A = 24.20 cm^3 + 24.10 cm^32 V_A = 48.30 cm^32 V_A = 24.15 cm^3 (ii) Concentration of E in moldm^-3 Step 1: Use the titration formula (C_A V_A)/(C_B V_B) = (n_A)/(n_B) to find the concentration of the base (C_B). C_B = (n_B C_A V_A)/(n_A V_B) Step 2: Substitute the known values. C_B = 2 × 0.05 moldm^-3 × 24.15 cm^31 × 25.0 cm^3 C_B = 2.415 moldm^-325.0 C_B = 0.0966 moldm^-3 (iii) Concentration of E in gdm^-3 Step 1: Calculate the molar mass of KOH. Molar mass of KOH = K + O + H = 39 + 16 + 1 = 56 g/mol. Step 2: Convert the concentration from moldm^-3 to gdm^-3. Concentration in gdm^-3 = Concentration in moldm^-3 × Molar mass Concentration of pure KOH = 0.0966 moldm^-3 × 56 g/mol Concentration of pure KOH = 5.41 gdm^-3 (iv) Percentage of impurity in an impure solution E Step 1: Calculate the mass of impurity. Mass of impurity = Concentration of impure KOH - Concentration of pure KOH Mass of impurity = 7.50 gdm^-3 - 5.41 gdm^-3 Mass of impurity = 2.09 gdm^-3 Step 2: Calculate the percentage of impurity. Percentage impurity = Mass of impurityConcentration of impure KOH × 100\% Percentage impurity = 2.09 gdm^-37.50 gdm^-3 × 100\% Percentage impurity = 0.27866 × 100\% Percentage impurity = 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 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. Send me the next one 📸