Two solutions, A and B, each of 100L was made by dissolving 4g of NaOH and 9.8 g of H₂SO₄ in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is :

In 100 L solution H₂SO₄ present = 9.8 gm <br><br>$\therefore$ In 10 L solution H₂SO₄ present = $9.8 \times {{10} \over {100}}$ gm <br><br>$\Rightarrow$ In 10 L solution moles of H₂SO₄ present = ${{9.8} \over {98}} \times {{10} \over {100}}$ <br><br>In one molecule of H₂SO₄ two H⁺ ion present. <br><br>$\therefore$ In 10 L solution moles of H⁺ present = 2$\times$${{9.8} \over {98}} \times {{10} \over {100}}$ = 0.02 moles <br><br>Also In 100 L solution NaOH present = 4 gm <br><br>$\therefore$ In 40 L solution NaOH present = $4 \times {{40} \over {100}}$ <br><br>$\Rightarrow$ In 40 L solution moles of NaOH present = ${4 \over {40}} \times {{40} \over {100}}$ <br><br>In one molecule of NaOH one OH^- ion present. <br><br>$\therefore$ In 40 L solution moles of OH^- ion present = ${4 \over {40}} \times {{40} \over {100}}$ = 0.04 moles <br><br>As moles of OH^- ion is more than H⁺ ion, so solution is basic. <br><br>$\therefore$ Final Conc. of OH^– = ${{0.04 - 0.02} \over {40 + 10}}$ = 4 $\times$ 10^-4 <br><br>$\therefore$ pOH = – log (4 ×10^–4) = 3.4 <br><br>pH = 14 – 3.4 = 10.6