JEE MAIN - Chemistry (2003 - No. 54)
25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was
0.14
0.28
0.35
0.07
Explanation
Ba(OH)2 + 2HCl(aq) $$ \to $$ BaCl2(aq) + 2H2O(l)
Assume molarity of Ba(OH)2 = M
No of moles of Ba(OH)2 present = 25$$\times$$M
No of moles of HCl present = 35$$\times$$0.1
Stoichiometric coefficient of Ba(OH)2 = 1
Stoichiometric coefficient of HCl = 2
$$\therefore$$ $${{25 \times M} \over 1}$$ = $${{35 \times 0.1} \over 2}$$
$$ \Rightarrow $$ M = 0.07
Assume molarity of Ba(OH)2 = M
No of moles of Ba(OH)2 present = 25$$\times$$M
No of moles of HCl present = 35$$\times$$0.1
Stoichiometric coefficient of Ba(OH)2 = 1
Stoichiometric coefficient of HCl = 2
$$\therefore$$ $${{25 \times M} \over 1}$$ = $${{35 \times 0.1} \over 2}$$
$$ \Rightarrow $$ M = 0.07
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