Please chemistry

• Given:

  • Acid solution A: $\text{KHSO}_4$
  • Base solution B: $\text{KOH}$
  • Volume of acid A ($V_A$) = $22.60 cm^3$
  • Volume of base B ($V_B$) = $25.0 cm^3$
  • Concentration of A = $12.0 \text{ g}$ of $\text{KHSO}_4$ per $\text{dm}^3$
  • Concentration of B = $Y \text{ g}$ of $\text{KOH}$ per $\text{dm}^3$
  • Equation: $\text{KHSO}_{4(aq)} + KOH_{(aq)} \longrightarrow K_2SO_{4(aq)} + H_2O_{(l)}$
  • Atomic masses: $\text{K}=39, H=1, S=32, O=16$

• Molar masses:

  • $M(KHSO_4) = 39 + 1 + 32 + (4 \times 16) = 136 \text{ g/mol}$
  • $M(KOH) = 39 + 16 + 1 = 56 \text{ g/mol}$
  • $M(K_2SO_4) = (2 \times 39) + 32 + (4 \times 16) = 78 + 32 + 64 = 174 \text{ g/mol}$

(a) From the information given above calculate the:

(i) Concentration of A in $\text{moldm}^{-3}$ Step 1: Convert the mass concentration of A to molar concentration. $C_A = \frac{Mass concentration}{Molar mass of KHSO_4}$ $C_A = \frac{12.0 gdm^{-3}}{136 g/mol}$ C_A = \text{0.0882 moldm^{-3}}

(ii) Concentration of B in $\text{moldm}^{-3}$ Step 1: Determine the mole ratio from the balanced equation. From $\text{KHSO}_{4(aq)} + KOH_{(aq)} \longrightarrow K_2SO_{4(aq)} + H_2O_{(l)}$, the mole ratio of acid to base ($n_A : n_B$) is $1:1$. Step 2: Use the titration formula $\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$ to find $C_B$. $C_B = \frac{n_B C_A V_A}{n_A V_B}$ Step 3: Substitute the known values. $C_B = \frac{1 \times 0.088235 moldm^{-3} \times 22.60 cm^3}{1 \times 25.0 cm^3}$ $C_B = \frac{1.994051}{25.0} moldm^{-3}$ C_B = \text{0.0798 moldm^{-3}}

(iii) Value of Y. Step 1: Y is the concentration of $\text{KOH}$ in $\text{gdm}^{-3}$. Convert the molar concentration of B to mass concentration. $Y = C_B \times M(KOH)$ $Y = 0.079762 moldm^{-3} \times 56 \text{ g/mol}$ Y = \text{4.47 gdm^{-3}}

(iv) Mass of $\text{K}_2SO_4$ formed during the reaction. Step 1: Calculate the moles of $\text{KHSO}_4$ reacted. $n_A = C_A \times V_A$ $n_A = 0.088235 moldm^{-3} \times \frac{22.60}{1000} dm^3$