JEE MAIN - Chemistry (2019 - 11th January Evening Slot - No. 10)
25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solutions ?
50 mL
12.5 mL
25 mL
75 mL
Explanation
2HCl(aq) + Na2CO3(aq) $$ \to $$ H2CO3 + NaCl
$$ \Rightarrow $$ $${{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}$$
$$ \Rightarrow $$ Molarity of HCl (M) = $${6 \over {25}}M$$
HCl(aq) + NaOH(aq) $$ \to $$ NaCl + H2O
$${{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}$$
$$ \Rightarrow $$ V = 25 ml
moles of HCl
2
=
moles of Na2CO3
1
$$ \Rightarrow $$ $${{M \times {{25} \over {1000}}} \over 2} = {{0.1 \times {{30} \over {1000}}} \over 1}$$
$$ \Rightarrow $$ Molarity of HCl (M) = $${6 \over {25}}M$$
HCl(aq) + NaOH(aq) $$ \to $$ NaCl + H2O
moles of HCl
1
=
moles of NaOH
1
$${{{6 \over {25}} \times {V \over {1000}}} \over 1} = {{0.2 \times {{30} \over {1000}}} \over 1}$$
$$ \Rightarrow $$ V = 25 ml
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