10 mL of 1 M H_2SO_4 will completely neutra | vedclass

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(B) The neutralization reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. Number of millimoles of $H_2SO_4 = M 	imes V 	ext{ (in mL)} = 1

Vedclass · Accepted Answer

$10 \ mL$ of $2 \ M$ $NaOH$ solution

Vedclass · Answer

(B) The neutralization reaction is: $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. Number of millimoles of $H_2SO_4 = M 	imes V 	ext{ (in mL)} = 1 	imes 10 = 10 \ mmol$. Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$,$10 \ mmol$ of $H_2SO_4$ requires $20 \ mmol$ of $NaOH$. Checking the options: For option $(B)$: $M 	imes V = 2 \ M 	imes 10 \ mL = 20 \ mmol$ of $NaOH$. Thus,$10 \ mL$ of $2 \ M$ $NaOH$ is required for complete neutralization.

$10 \ mL$ of $1 \ M$ $H_2SO_4$ will completely neutralize:

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