JEE MAIN - Chemistry (2020 - 7th January Morning Slot - No. 1)
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 :
Answer
10.6
Explanation
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
$$ \therefore $$ 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
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