Calculate [H^{+}] at the equivalence point du | vedclass

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(B) At the equivalence point,the moles of $HA$ equal the moles of $NaOH$ added.Moles of $HA = 0.1 \ M 	imes 0.025 \ L = 0.0025 \ mol$.Volume of $NaOH

Vedclass · Accepted Answer

$1.732 	imes 10^{-9} \ M$

Vedclass · Answer

(B) At the equivalence point,the moles of $HA$ equal the moles of $NaOH$ added.Moles of $HA = 0.1 \ M 	imes 0.025 \ L = 0.0025 \ mol$.Volume of $NaOH$ required $= \frac{0.0025 \ mol}{0.05 \ M} = 0.05 \ L = 50 \ mL$.Total volume of the solution $= 25 \ mL + 50 \ mL = 75 \ mL = 0.075 \ L$.Concentration of salt $NaA$ formed $= \frac{0.0025 \ mol}{0.075 \ L} = \frac{1}{30} \ M$.For the salt of a weak acid and strong base,$[OH^{-}] = \sqrt{K_h 	imes C} = \sqrt{\frac{K_w}{K_a} 	imes C}$.$[H^{+}] = \frac{K_w}{[OH^{-}]} = \sqrt{\frac{K_w 	imes K_a}{C}}$.Substituting the values: $[H^{+}] = \sqrt{\frac{10^{-14} 	imes 10^{-5}}{1/30}} = \sqrt{30 	imes 10^{-19}} = \sqrt{3 	imes 10^{-18}} = 1.732 	imes 10^{-9} \ M$.

Calculate $[H^{+}]$ at the equivalence point during the titration of $25 \ mL$ of $0.1 \ M$ weak acid $HA$ $(K_a = 10^{-5})$ with $0.05 \ M \ NaOH$ solution.

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