Question

The volume of ${\text{0}}{\text{.1 N}}$ dibasic acid sufficient to neutralize ${\text{1 g}}$ of a base that furnishes ${\text{0}}{\text{.04 mol}}$ of ${\text{O}}{{\text{H}}^ - }$ in aqueous solution is:
A) ${\text{400 ml}}$
B) ${\text{600 ml}}$
C) ${\text{200 ml}}$
D) ${\text{800 ml}}$

Answer 1

We will use the following relation between the number of gram equivalents of acid and the number of gram equivalents of hydroxide ions.
${\text{number of gram equivalents of acid = number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}}$
We will also use the following relationship between the number of gram equivalents of acid, normality and volume.
Here, ${\text{number of gram equivalents of acid = normality }} \times {\text{ volume}}$

Complete step-by-step solution:
Hydroxide ion is a monoacidic base. For hydroxide ions, the number of gram equivalents is equal to the number of moles. This is because the gram equivalent weight of hydroxide ion is equal to the molecular weight of hydroxide ion. Gram equivalent weight of a base is equal to the ratio of its molecular weight to its acidity. The acidity of hydroxide ions is one. We will first obtain the number of gram equivalents of hydroxide ions from the number of moles of hydroxide ions.

$${\text{number of moles of O}}{{\text{H}}^ - }{\text{ ions}} = {\text{ 0}}{\text{.04 mol}} \\\ {\text{number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}} = {\text{number of moles of O}}{{\text{H}}^ - }{\text{ ions}} \\\ {\text{number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}} = {\text{0}}{\text{.04 gram eq}} \\\$$

The normality of dibasic acid is ${\text{0}}{\text{.1 N}}$ . Now, we will calculate the volume of dibasic acid needed

$${\text{number of gram equivalents of acid = number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}} \\\ {\text{number of gram equivalents of acid = normality of acid}} \times {\text{ volume of acid}} \\\ {\text{normality of acid}} \times {\text{ volume of acid}} = {\text{number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}} \\\$$

We will substitute the values in the above equation and calculate the volume of dibasic acid needed.

$${\text{normality of acid}} \times {\text{ volume of acid}} = {\text{number of gram equivalents of O}}{{\text{H}}^ - }{\text{ ions}} \\\ {\text{0}}{\text{.1 N}} \times {\text{ volume of acid}} = {\text{0}}{\text{.04 gram eq}} \\\ {\text{ volume of acid}} = \dfrac{{{\text{0}}{\text{.04 gram eq}}}}{{{\text{0}}{\text{.1 N}}}} \\\ {\text{ volume of acid}} = 0.400{\text{ L}} \\\$$

We will convert the unit of volume from liter to milliliter.

$${\text{volume of acid}} = 0.400{\text{ L}} \\\ {\text{ volume of acid}} = 0.400{\text{ L}} \times {\text{ 1000 ml/L}} \\\ {\text{volume of acid}} = 400{\text{ mL}} \\\$$

Thus, the volume of acid needed is ${\text{400 ml}}$

Thus the correct answer is option ‘A’.

Note: Normality is the number of gram equivalents of solute present in one liter of solution. The number of gram equivalents is the ratio of the mass to the gram equivalent weight. The gram equivalent weight is equal to the ratio of the molecular weight to the acidity or basicity. Acidity represents the number of hydroxide ions obtained through dissociation of one molecule of base.