What volume of 0.1 , M , H_2SO_4 is required | vedclass

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(C) The neutralization reaction is: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$. Using the molarity equation for neutralization: $M_1V_1n_1 = M_2V_

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The neutralization reaction is: $2NaOH + H_2SO_4 ightarrow Na_2SO_4 + 2H_2O$. Using the molarity equation for neutralization: $M_1V_1n_1 = M_2V_2n_2$,where $n$ is the acidity/basicity factor. For $NaOH$,$n_1 = 1$ and for $H_2SO_4$,$n_2 = 2$. Given: $M_1 = 0.2 \, M$,$V_1 = 40 \, mL$,$M_2 = 0.1 \, M$. Substituting the values: $0.2 	imes 40 	imes 1 = 0.1 	imes V_2 	imes 2$. $8 = 0.2 	imes V_2$. $V_2 = \frac{8}{0.2} = 40 \, mL$.

What volume of $0.1 \, M \, H_2SO_4$ is required to completely neutralize $40 \, mL$ of $0.2 \, M \, NaOH$?

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