10 ,mL of a solution of NaOH is found to be c | vedclass

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Question

(A) According to the law of stoichiometry,the neutralization reaction between $NaOH$ and $HCl$ occurs in a fixed ratio.Given that $10 \,mL$ of $NaOH$

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) According to the law of stoichiometry,the neutralization reaction between $NaOH$ and $HCl$ occurs in a fixed ratio.Given that $10 \,mL$ of $NaOH$ requires $8 \,mL$ of $HCl$ for complete neutralization.If the volume of $NaOH$ is doubled to $20 \,mL$,the amount of $HCl$ required will also be doubled proportionally.Calculation: $(20 \,mL \,NaOH / 10 \,mL \,NaOH) 	imes 8 \,mL \,HCl = 2 	imes 8 \,mL = 16 \,mL$.Therefore,$16 \,mL$ of $HCl$ solution is required.

$10 \,mL$ of a solution of $NaOH$ is found to be completely neutralized by $8 \,mL$ of a given solution of $HCl$. If we take $20 \,mL$ of the same solution of $NaOH$,the amount of $HCl$ solution (the same solution as before) required to neutralize it will be ...... $mL$.

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