For which of the following solution, minimum amount of NaOH is required for complete neutralisation. [Assume complete dissociation of acid ]

Engineering

Chemistry

Concentration Terms

Question

For which of the following solution, minimum amount of NaOH is required for complete neutralisation.
[Assume complete dissociation of acid ]

100 mL HNO₃ solution [basicity = 1] (d_solution = 1.5 g/mL) , % w/w = 36.5

50 mL H₃PO₄ solution [basicity = 3] (d_solution = 2 g/mL) , % w/w = 98

50 mL H₂SO₄ solution [basicity = 2] (d_solution = 2 g/mL) , % w/w = 98

100 mL HCl solution [basicity = 1] (d_solution = 1.5 g/mL) , % w/w = 36.5

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(A) $M_{H_{2} {SO}_{4}} = \frac{10 \times 98 \times 2}{98} = 20$
       H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
       moles of H₂SO₄ = 20 × $\frac{50}{1000}$ = 1
       moles of NaOH = 2 × moles of H₂SO₄ = 2
(B) $M_{H_{3} {PO}_{4}} = \frac{10 \times 98 \times 2}{98} = 20$
       H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
       moles of H₃PO₄ = 20 × $\frac{50}{1000}$ = 1
       moles of NaOH = 3 × 1 = 3
(C)   M_HCl = $\frac{10 \times 36 .5 \times 2}{36 .5}$ = 20
       moles of NaOH = moles of HCl = 20 × $\frac{100}{1000}$ = 2
(D)    $M_{{HNO}_{3}} = \frac{10 \times 36 .5 \times 1 .5}{63}$
       moles of NaOH = moles of HNO₃ = $\frac{10 \times 36 .5 \times 1 .5}{63} \times \frac{100}{1000}$

$= \frac{36 .5 \times 1 .5}{63} 0 .87$