JEE MAIN - Chemistry (2004 - No. 53)
To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is
40 mL
20 mL
10 mL
60 mL
Explanation
H3PO3 + 2KOH $$ \to $$ K2HPO3 + 2H2O
Let the volume of KOH solution is = V mL = $${V \over {1000}}$$ litre
Volume of H3PO3 = 20 mL = $${{20} \over {1000}}$$ litre
No of moles of H3PO3 present = $${{20} \over {1000}} \times 0.1$$
No of moles of KOH present = $${V \over {1000}} \times 0.1$$
Stoichiometric coefficient of H3PO3 = 1
Stoichiometric coefficient of KOH = 2
$$\therefore$$ $${{{{20} \over {1000}} \times 0.1} \over 1}$$ = $${{{V \over {1000}} \times 0.1} \over 2}$$
$$ \Rightarrow $$ V = 40 mL
Let the volume of KOH solution is = V mL = $${V \over {1000}}$$ litre
Volume of H3PO3 = 20 mL = $${{20} \over {1000}}$$ litre
No of moles of H3PO3 present = $${{20} \over {1000}} \times 0.1$$
No of moles of KOH present = $${V \over {1000}} \times 0.1$$
Stoichiometric coefficient of H3PO3 = 1
Stoichiometric coefficient of KOH = 2
$$\therefore$$ $${{{{20} \over {1000}} \times 0.1} \over 1}$$ = $${{{V \over {1000}} \times 0.1} \over 2}$$
$$ \Rightarrow $$ V = 40 mL
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