If 25 mL of H_2SO_4 solution is neutralized | vedclass

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(A) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$. Using the titration formula $n_1 M_1 V_1 = n_2 M_2 V_2$,where $n$

Vedclass · Accepted Answer

$0.107$

Vedclass · Answer

(A) The neutralization reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. Using the titration formula $n_1 M_1 V_1 = n_2 M_2 V_2$,where $n$ is the acidity/basicity factor. For $H_2SO_4$,$n_1 = 2$ and for $NaOH$,$n_2 = 1$. Substituting the values: $2 	imes M_1 	imes 25 \ mL = 1 	imes 0.164 \ M 	imes 32.63 \ mL$. $M_1 = \frac{0.164 	imes 32.63}{2 	imes 25} = \frac{5.35132}{50} = 0.107 \ M$.

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

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