QJEEM20Che63e: JEE Main 2020 Chemistry - Class 11 JEEM20Che63e Solution: Two solutions, A and B, each of 100L was made by dissolving 4g…

In 100 L solution H₂SO₄ present = 9.8 gm

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

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

In one molecule of H₂SO₄ two H⁺ ion present.

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

Also In 100 L solution NaOH present = 4 gm

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

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

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

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

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

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

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

pH = 14 – 3.4 = 10.6