If 25 mL of a H_2SO_4 solution is neutralize | vedclass

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Question

(A) The balanced chemical equation for the neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. From the stoichiometry,$1 \ mol

Vedclass · Accepted Answer

$0.107$

Vedclass · Answer

(A) The balanced chemical equation for the neutralization reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. From the stoichiometry,$1 \ mole$ of $H_2SO_4$ reacts with $2 \ moles$ of $NaOH$. Using the titration formula $M_1V_1 / n_1 = M_2V_2 / n_2$,where $M_1, V_1$ are for $H_2SO_4$ and $M_2, V_2$ are for $NaOH$: $M_1 	imes 25 / 1 = 0.164 	imes 32.63 / 2$. $M_1 = (0.164 	imes 32.63) / (2 	imes 25)$. $M_1 = 5.35132 / 50 = 0.1070264 \ M$. Rounding to three decimal places,the molarity is $0.107 \ M$.

If $25 \ mL$ of a $H_2SO_4$ solution is neutralized by $32.63 \ mL$ of $0.164 \ M \ NaOH$ solution,what is the molarity of the $H_2SO_4$ solution (in $M$)?

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