(D) At the equivalence point,the number of millimoles of the acid $HA$ equals the number of millimoles of the base $NaOH$. Since $HA$ is a monoprotic

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(D) At the equivalence point,the number of millimoles of the acid $HA$ equals the number of millimoles of the base $NaOH$. Since $HA$ is a monoprotic acid,the millimoles of $H^+$ equal the millimoles of $HA$. $\text{Millimoles of } HA = M \times V = 0.1 \ M \times 20 \ mL = 2 \ \text{mmol}$. $\text{Millimoles of } NaOH = M \times V = 0.25 \ M \times V_{NaOH}$. Equating the two: $2 = 0.25 \times V_{NaOH}$. $V_{NaOH} = 2 / 0.25 = 8 \ mL$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Mix Examples-Ionic Equilibrium

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$A$ $20 \ mL$ sample of a $0.1 \ M$ solution of the weak monoprotic acid $HA$ is titrated with $0.25 \ M \ NaOH$. The volume of $NaOH$ solution required to reach the equivalence point is: ............. $mL$

A
$10$
B
$6$
C
$12$
D
$8$

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2 Equilibrium: Ionic Equilibrium

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Mix Examples-Ionic Equilibrium

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If $pK_{a} = -\log K_{a} = 4$ for a weak acid $HX$ and $K_{a} = C\alpha^{2}$,then the Van't Hoff factor when $C = 0.01 \ M$ is

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$A$ $20 \ mL$ sample of a $0.1 \ M$ solution of the weak monoprotic acid $HA$ is titrated with $0.25 \ M \ NaOH$. The volume of $NaOH$ solution required to reach the equivalence point is: ............. $mL$

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