JEE Main 2020 — Ionic Equilibrium Question with Solution

In 100 L solution H₂SO₄ present = 9.8 gm

$$\therefore$$ In 10 L solution H₂SO₄ present = $$9.8\times \frac{10}{100}$$ gm

$$\Rightarrow$$ In 10 L solution moles of H₂SO₄ present = $$\frac{9.8}{98}\times \frac{10}{100}$$

In one molecule of H₂SO₄ two H⁺ ion present.

$$\therefore$$ In 10 L solution moles of H⁺ present = 2$$\times$$$$\frac{9.8}{98}\times \frac{10}{100}$$ = 0.02 moles

Also In 100 L solution NaOH present = 4 gm

$$\therefore$$ In 40 L solution NaOH present = $$4\times \frac{40}{100}$$

$$\Rightarrow$$ In 40 L solution moles of NaOH present = $$\frac{4}{40}\times \frac{40}{100}$$

In one molecule of NaOH one OH^- ion present.

$$\therefore$$ In 40 L solution moles of OH^- ion present = $$\frac{4}{40}\times \frac{40}{100}$$ = 0.04 moles

As moles of OH^- ion is more than H⁺ ion, so solution is basic.

$$\therefore$$ Final Conc. of OH^– = $$\frac{0.04-0.02}{40+10}$$ = 4 $$\times$$ 10^-4

$$\therefore$$ pOH = – log (4 ×10^–4) = 3.4

pH = 14 – 3.4 = 10.6