2.5 mL of 0.4 M weak monoacidic base (K_{b} | vedclass

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Question

(C) At equivalence point,the moles of base equal the moles of acid: $n_{base} = n_{acid}$.$0.4 \ M 	imes 2.5 \ mL = \frac{2}{15} \ M 	imes V_{acid}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) At equivalence point,the moles of base equal the moles of acid: $n_{base} = n_{acid}$.$0.4 \ M 	imes 2.5 \ mL = \frac{2}{15} \ M 	imes V_{acid}$.$V_{acid} = \frac{0.4 	imes 2.5 	imes 15}{2} = 7.5 \ mL$.Total volume $= 2.5 \ mL + 7.5 \ mL = 10 \ mL$.Concentration of salt $(C)$ $= \frac{0.4 	imes 2.5}{10} = 0.1 \ M$.The salt formed is a salt of a weak base and a strong acid,which undergoes cationic hydrolysis.The formula for $[H^{+}]$ is $[H^{+}] = \sqrt{\frac{K_{w} 	imes C}{K_{b}}}$.$[H^{+}] = \sqrt{\frac{10^{-14} 	imes 0.1}{10^{-12}}} = \sqrt{10^{-3}} = \sqrt{10 	imes 10^{-4}} = 3.16 	imes 10^{-2} \ M \approx 3.2 	imes 10^{-2} \ M$.

$2.5 \ mL$ of $0.4 \ M$ 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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