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(D) For the $2^{nd}$ equivalence point,the reaction is: $H_2A + 2NaOH \rightarrow Na_2A + 2H_2O$ From the stoichiometry of the reaction: $n_{NaOH} = 2

Vedclass · Accepted Answer

$0.0769 \ M \ Na_2A$

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(D) For the $2^{nd}$ equivalence point,the reaction is: $H_2A + 2NaOH ightarrow Na_2A + 2H_2O$ From the stoichiometry of the reaction: $n_{NaOH} = 2 	imes n_{H_2A}$ $0.25 \ M 	imes V_{NaOH} = 2 	imes (20 \ mL 	imes 0.2 \ M)$ $0.25 	imes V_{NaOH} = 8 	ext{ mmol}$ $V_{NaOH} = 32 \ mL$ Total volume of the solution = $20 \ mL + 32 \ mL = 52 \ mL$ At the $2^{nd}$ equivalence point,all $H_2A$ is converted to $Na_2A$. $n_{Na_2A} = n_{H_2A} = 20 \ mL 	imes 0.2 \ M = 4 	ext{ mmol}$ Concentration of $Na_2A = \frac{n_{Na_2A}}{V_{total}} = \frac{4 	ext{ mmol}}{52 \ mL} \approx 0.0769 \ M$

$A$ $20 \ mL$ sample of a $0.2 \ M$ solution of the weak diprotic acid $H_2A$ is titrated with $0.25 \ M \ NaOH$. Molarity of the solution at the second equivalence point is:

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