The density of a monobasic strong acid (Molar | vedclass

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Question

(B) For a monobasic acid,the reaction with $NaOH$ is: $HA + NaOH \rightarrow NaA + H_2O$.Since the acid is monobasic,$1 \ mol$ of acid reacts with $1

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) For a monobasic acid,the reaction with $NaOH$ is: $HA + NaOH ightarrow NaA + H_2O$.Since the acid is monobasic,$1 \ mol$ of acid reacts with $1 \ mol$ of $NaOH$.Millimoles of $NaOH = M 	imes V(mL) = 0.24 	imes 25 = 6 \ mmol$.Therefore,millimoles of acid required $= 6 \ mmol$.Mass of acid $= 	ext{moles} 	imes 	ext{molar mass} = 6 	imes 10^{-3} \ mol 	imes 24.2 \ g \ mol^{-1} = 0.1452 \ g$.Given density $d = 1.21 \ kg \ L^{-1} = 1.21 \ g \ mL^{-1}$.Volume of acid solution $V = \frac{	ext{mass}}{	ext{density}} = \frac{0.1452 \ g}{1.21 \ g \ mL^{-1}} = 0.12 \ mL$.Converting to the required format: $0.12 \ mL = 12 	imes 10^{-2} \ mL$.

The density of a monobasic strong acid (Molar mass $24.2 \ g \ mol^{-1}$) is $1.21 \ kg \ L^{-1}$. The volume of its solution required for the complete neutralization of $25 \ mL$ of $0.24 \ M \ NaOH$ is $.............. \times 10^{-2} \ mL$ (Nearest integer).

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