The volume of 0.6 , M NaOH required to neutra | vedclass

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(C) The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.According to the law of equivalence,the number of moles of $H^+$ must equal t

Vedclass · Accepted Answer

$20$

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(C) The neutralization reaction is: $NaOH + HCl ightarrow NaCl + H_2O$.According to the law of equivalence,the number of moles of $H^+$ must equal the number of moles of $OH^-$.For $HCl$,the number of moles $= Molarity 	imes Volume = 0.4 \, M 	imes 30 \, cm^3 = 12 \, mmol$.Since $NaOH$ is a monobasic base,$1 \, mol$ of $NaOH$ reacts with $1 \, mol$ of $HCl$.Therefore,$12 \, mmol$ of $NaOH$ is required.$Volume \, of \, NaOH = \frac{Moles}{Molarity} = \frac{12 \, mmol}{0.6 \, M} = 20 \, cm^3$.

The volume of $0.6 \, M$ $NaOH$ required to neutralize $30 \, cm^3$ of $0.4 \, M$ $HCl$ is ........ $cm^3$.

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