Question

Find concentration of [H+] after mixing 15 mL, 0.1M H2SO4 and 15 mL 0.1M NaOH :-
(i) 5 × 10^-2
(ii) 10^-7
(iii) 1 × 10^-2
(iv) 2 × 10^-2​

Answer 1

Given :

• Volume of H₂SO₄ = 15ml
• Molarity of H₂SO₄ = 0.1 M
• Volume of NaOH = 15 ml
• Molarity of NaOH = 0.1 M

To find :

Concentration of [H⁺] ion after mixing given Solution

Solution :

H₂SO₄ is strong acid and NaOH is strong base, therefore on mixing neutralization reaction will occur.

H₂SO₄ + 2NaOH ➝ Na₂SO₄ + 2H₂O

________________________________

Now we will find number of moles of each reactant and hence limiting reactant.

Note -

[ Formula used - Number of mole = Molarity × volume in L ]

1) Moles of H₂SO₄ = molarity of H₂SO₄ × volume of H₂SO₄

➝ moles of H₂SO₄ = 0.1 × (15/1000)

➝ moles of H₂SO₄ = 1.5 /1000 mol

➝ moles of H₂SO₄ = 1.5 milimole

2) Moles of NaOH = molarity of NaOH × volume of NaOH

➝ moles of NaOH = 0.1 × (15/1000)

➝ moles of NaOH = 1.5 /1000 mol

➝ moles of NaOH = 1.5 milimole

________________________________

Now , according to balanced chemical equation

➝ 2 mole of NaOH require 1 mole of H₂SO₄ for neutralization

➝ 1 mole of NaOH require 1/2 mole of H₂SO₄ for neutralization

➝ 1.5 milimoles of NaOH will require 1.5/2 milimole of H₂SO₄ for neutralization

➝ 1.5 milimoles of NaOH will require 0.75 milimole of H₂SO₄ for neutralization

From above it is clear that NaOH is limiting reactant and will be utilised completely. And after completion of reaction only H₂SO₄ will remain in Solution

________________________________

Amount of H₂SO₄ left = Initial amount - amount utilised in reaction

➝ Amount of H₂SO₄ left = 1.5 milimoles - 0.75 milimoles

➝ Amount of H₂SO₄ left = 0.75 milimoles

________________________________

H₂SO₄ ➝ 2H⁺ + SO₄²⁻

From this reaction it is clear that

1 mole of H₂SO₄ will give 2 mole of H⁺ ion

➝ 0.75 milimole of H₂SO₄ will give 2×0.75 milimole of H⁺ ion

➝ 0.75 milimole of H₂SO₄ will give 1.5 milimole of H⁺ ion

________________________________

Therefore.

• Amount of H⁺ ion Present in solution = 1.5 milimole
• Final volume of Solution = 15ml + 15ml = 30ml

Concentration of H⁺ ion = Molarity of H⁺ ion

➝ Concentration of H⁺ ion = Number of mole of H⁺ ion ÷ Volume of Solution in L

$\sf \implies \:Concentration \: of \: H^+ \: ion = \: \dfrac{1.5 \times {10}^{ - 3} }{30 \times {10}^{ - 3} } \: M\\ \\ \sf \implies \:Concentration \: of \: H^+ \: ion = \: \dfrac{1.5 }{30 } \: M \\ \\ \sf \implies \:Concentration \: of \: H^+ \: ion = \: \dfrac{15 }{300 } \: M \\ \\ \sf \implies \:Concentration \: of \: H^+ \: ion = \: \dfrac{15 }{3 } \: \times {10}^{ - 2} \: M \\ \\ \sf \implies \:Concentration \: of \: H^+ \: ion = \: \bold{ 5 \: \times {10}^{ - 2} \: M}$

________________________________

ANSWER:

Option (i) 5 × 10⁻² M