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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Given: Acid solution A: KHSO_4 Base solution B: KOH Volume of acid A (V_A) = 22.60 cm^3 Volume of base B (V_B) = 25.0 cm^3 Concentration of A = 12.0 g of KHSO_4 per dm^3 Concentration of B = Y g of KOH per dm^3 Equation: KHSO_4(aq) + KOH_(aq) K_2SO_4(aq) + H_2O_(l) Atomic masses: K=39, H=1, S=32, O=16 Molar masses: M(KHSO_4) = 39 + 1 + 32 + (4 × 16) = 136 g/mol M(KOH) = 39 + 16 + 1 = 56 g/mol M(K_2SO_4) = (2 × 39) + 32 + (4 × 16) = 78 + 32 + 64 = 174 g/mol (a) From the information given above calculate the: (i) Concentration of A in moldm^-3 Step 1: Convert the mass concentration of A to molar concentration. C_A = Mass concentrationMolar mass of KHSO_4 C_A = 12.0 gdm^-3136 g/mol C_A = 0.0882 moldm^-3 (ii) Concentration of B in moldm^-3 Step 1: Determine the mole ratio from the balanced equation. From KHSO_4(aq) + KOH_(aq) K_2SO_4(aq) + H_2O_(l), the mole ratio of acid to base (n_A : n_B) is 1:1. Step 2: Use the titration formula (C_A V_A)/(C_B V_B) = (n_A)/(n_B) to find C_B. C_B = (n_B C_A V_A)/(n_A V_B) Step 3: Substitute the known values. C_B = 1 × 0.088235 moldm^-3 × 22.60 cm^31 × 25.0 cm^3 C_B = (1.994051)/(25.0) moldm^-3 C_B = 0.0798 moldm^-3 (iii) Value of Y. Step 1: Y is the concentration of KOH in gdm^-3. Convert the molar concentration of B to mass concentration. Y = C_B × M(KOH) Y = 0.079762 moldm^-3 × 56 g/mol Y = 4.47 gdm^-3 (iv) Mass of K_2SO_4 formed during the reaction. Step 1: Calculate the moles of KHSO_4 reacted. n_A = C_A × V_A n_A = 0.088235 moldm^-3 × (22.60)/(1000) dm^3 n_A = 0.001994051 mol