Chemistry - Concentration Terms Question with Solution | TestHub

Solution: The reaction is P₄O₁₀ + 6H₂O -> 4H₃PO₄.

Molar mass of P₄O₁₀ = $4\times 31+10\times 16=124+160=284$ g/mol.

Moles of P₄O₁₀ = $\frac{5.68g}{284g/mol}=0.02$ mol.

Moles of H₃PO₄ produced = $4\times 0.02=0.08$ mol. Volume of solution = 250 mL = 0.25 L.

Molarity of H₃PO₄ = $\frac{0.08mol}{0.25L}=0.32$ M. (Statement A is correct)

For the first equivalence point of H₃PO₄ with KOH, the reaction is H₃PO₄ + KOH -> KH₂PO₄ + H₂O. The n-factor for H₃PO₄ is 1.

Moles of H₃PO₄ in 40 mL = $0.32M\times 0.040L=0.0128$ mol.

Moles of KOH required = $0.0128$ mol.

Given KOH solution: $48mL\times 0.8N=0.048L\times 0.8mol/L=0.0384$ mol.

Since $0.0128\ne 0.0384$, statement D is incorrect.

Explanation:

Statement A: Molarity of H₃PO₄ is calculated correctly as 0.32 M.

Statement B: For complete neutralization, H₃PO₄ + 3NaOH -> Na₃PO₄ + 3H₂O.

Moles of H₃PO₄ in 25 mL = $0.32M\times 0.025L=0.008$ mol.

Moles of NaOH required = $3\times 0.008=0.024$ mol.

Volume of 0.5 M NaOH required = $\frac{0.024mol}{0.5mol/L}=0.048L=48$ mL. (Statement B is correct)

Statement C: For complete precipitation of phosphate, 3BaCl₂ + 2H₃PO₄ -> Ba₃(PO₄)₂ + 6HCl.

Moles of H₃PO₄ in 15 mL = $0.32M\times 0.015L=0.0048$ mol.

Moles of BaCl₂ required = $\frac{3}{2}\times 0.0048=0.0072$ mol.

Volume of 0.2 M BaCl₂ required = $\frac{0.0072mol}{0.2mol/L}=0.036L=36$ mL. (Statement C is correct)

Statement D: At the first equivalence point, 1 mole of H₃PO₄ reacts with 1 mole of KOH. The calculated moles of KOH required (0.0128 mol) do not match the given moles of KOH (0.0384 mol). Therefore, statement D is incorrect.

The final answer is $\boxed{D}$