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

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

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

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

(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 = $\frac{10 \times 36 .5 \times 1 .5}{63} \times \frac{100}{1000}$

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

Concept Explanation: Neutralization and Equivalent Weight

To determine which solution requires the minimum amount of NaOH for complete neutralization, we need to calculate the number of equivalents of each acid. The solution with the least number of equivalents will require the least NaOH, since NaOH provides one equivalent per mole (as it is monobasic).

Key Formula: Number of equivalents of acid = (Mass of acid) / (Equivalent weight of acid)

Where Equivalent weight = Molar mass / Basicity

Mass of acid = Volume of solution × Density × (% w/w / 100)

Step-by-Step Calculation for Each Option:

Step 1: For 100 mL HNO₃ solution (Basicity=1, d=1.5 g/mL, %w/w=36.5)

Mass of solution = 100 mL × 1.5 g/mL = 150 g

Mass of HNO₃ = 150 g × (36.5/100) = 54.75 g

Molar mass of HNO₃ = 63 g/mol

Equivalent weight = 63 / 1 = 63 g/equiv

Number of equivalents = 54.75 / 63 ≈ 0.869 equiv

Step 2: For 100 mL HCl solution (Basicity=1, d=1.5 g/mL, %w/w=36.5)

Mass of solution = 100 mL × 1.5 g/mL = 150 g

Mass of HCl = 150 g × (36.5/100) = 54.75 g

Molar mass of HCl = 36.5 g/mol

Equivalent weight = 36.5 / 1 = 36.5 g/equiv

Number of equivalents = 54.75 / 36.5 ≈ 1.5 equiv

Step 3: For 50 mL H₂SO₄ solution (Basicity=2, d=2 g/mL, %w/w=98)

Mass of solution = 50 mL × 2 g/mL = 100 g

Mass of H₂SO₄ = 100 g × (98/100) = 98 g

Molar mass of H₂SO₄ = 98 g/mol

Equivalent weight = 98 / 2 = 49 g/equiv

Number of equivalents = 98 / 49 = 2 equiv

Step 4: For 50 mL H₃PO₄ solution (Basicity=3, d=2 g/mL, %w/w=98)

Mass of solution = 50 mL × 2 g/mL = 100 g

Mass of H₃PO₄ = 100 g × (98/100) = 98 g

Molar mass of H₃PO₄ = 98 g/mol

Equivalent weight = 98 / 3 ≈ 32.667 g/equiv

Number of equivalents = 98 / 32.667 ≈ 3 equiv

Final Answer: The HNO₃ solution requires the minimum number of equivalents (≈0.869), so it needs the least amount of NaOH for complete neutralization.

Important Formulae and Theory

Equivalent Weight of Acid: $\frac{Molar mass}{Basicity}$

Number of Equivalents: $\frac{Mass of acid}{Equivalent weight}$

Mass of Acid in Solution: Volume × Density × (%w/w / 100)

Neutralization Reaction: Acid + Base → Salt + Water. The equivalents of acid equal equivalents of base.

Detailed Solution

Concept Explanation: Neutralization and Equivalent Weight

Step-by-Step Calculation for Each Option:

Related Topics

Important Formulae and Theory

Detailed Solution

Concept Explanation: Neutralization and Equivalent Weight

Step-by-Step Calculation for Each Option:

Related Topics

Important Formulae and Theory

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5