2.5 mL of frac{2}{5} M weak monoacidic base | vedclass

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Question

(C) The reaction is $BOH + HCl \longrightarrow BCl + H_2O$.At the equivalence point,the salt $BCl$ is formed.The moles of $BOH = 2.5 \ mL 	imes 0.4 \

Vedclass · Accepted Answer

$3.2 	imes 10^{-2} \ M$

Vedclass · Answer

(C) The reaction is $BOH + HCl \longrightarrow BCl + H_2O$.At the equivalence point,the salt $BCl$ is formed.The moles of $BOH = 2.5 \ mL 	imes 0.4 \ M = 1.0 \ mmol$.The volume of $HCl$ required $= \frac{1.0 \ mmol}{2/15 \ M} = 7.5 \ mL$.Total volume at equivalence point $= 2.5 \ mL + 7.5 \ mL = 10 \ mL$.Concentration of salt $C = \frac{1.0 \ mmol}{10 \ mL} = 0.1 \ M$.For the salt of a weak base and strong acid,$[H^{+}] = \sqrt{\frac{K_W 	imes C}{K_b}}$.$[H^{+}] = \sqrt{\frac{10^{-14} 	imes 0.1}{10^{-12}}} = \sqrt{10^{-3}} = \sqrt{10 	imes 10^{-4}} \approx 3.16 	imes 10^{-2} \ M$.

$2.5 \ mL$ of $\frac{2}{5} \ M$ weak monoacidic base ($K_{b} = 1 \times 10^{-12}$ at $25^{\circ} C$) is titrated with $\frac{2}{15} \ M \ HCl$ in water at $25^{\circ} C$. The concentration of $H^{+}$ at equivalence point is ($K_W = 1 \times 10^{-14}$ at $25^{\circ} C$).

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