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Question: 125 \( {{mL}} \) of 63% (w/v) \( {{{H}}_{{2}}}{{{C}}_{{2}}}{{{O}}_{{4}}}{{.2}}{{{H}}_{{2}}}{{O}} \) ...

125 mL{{mL}} of 63% (w/v) H2C2O4.2H2O{{{H}}_{{2}}}{{{C}}_{{2}}}{{{O}}_{{4}}}{{.2}}{{{H}}_{{2}}}{{O}} solution is made to react with 125 mL{{mL}} of a 40% (w/v) NaOH{{NaOH}} solution. The resulting solution is:
(A) Neutral
(B) Acidic
(C) Strongly acidic
(D) Alkaline

Explanation

Solution

This problem can be solved from the knowledge of the different concentration units that are used to express the concentration of the solution. We shall substitute appropriate values in the equation given.
Formula: %(w/v)=weight of the solutevolume of the solution×100{{\% }}\left( {{{w/v}}} \right){{ = }}\dfrac{{{\text{weight of the solute}}}}{{{\text{volume of the solution}}}}{{ \times 100}}

Complete Stepwise Solution
The percentage composition of a solution is expressed as the number of grams of the solute that dissolves in 100 grams of the solvent. The mathematical expression of the percentage composition is:
%(w/v)=weight of the solutevolume of the solution×100{{\% }}\left( {{{w/v}}} \right){{ = }}\dfrac{{{\text{weight of the solute}}}}{{{\text{volume of the solution}}}}{{ \times 100}}
The concept of milliequivalent was created to account for the fact that when solutes dissolve in solvents to form the solution, then the number of the dispersed particles depends on the valance of the solutes.
1 milliequivalent is defined as mass×valencymolecular weight\dfrac{{{{mass \times valency}}}}{{{\text{molecular weight}}}}
As the weight of oxalic acid in 100 mL{{mL}} is 63 g as per the given equation, the weight in 125 mL{{mL}} will be:
63100×125\dfrac{{63}}{{100}} \times 125 , converting this weight by volume percentage into milliequivalents,
1 Milliequivalent of H2C2O4.2H2O{{{H}}_{{2}}}{{{C}}_{{2}}}{{{O}}_{{4}}}{{.2}}{{{H}}_{{2}}}{{O}} = 63100×125×valencymolecular weight\dfrac{{63}}{{100}} \times 125 \times \dfrac{{{{valency}}}}{{{\text{molecular weight}}}} = 63100×125×2126\dfrac{{63}}{{100}} \times 125 \times \dfrac{2}{{{{126}}}} = 125100\dfrac{{125}}{{100}}
Similarly,
The weight of sodium hydroxide in 100 mL{{mL}} is 40 g as per the given equation, the weight in 125 mL{{mL}} will be:
40100×125\dfrac{{40}}{{100}} \times 125 , converting this weight by volume percentage into milliequivalents,
1 Milliequivalent of sodium hydroxide = 40100×125×140\dfrac{{40}}{{100}} \times 125 \times \dfrac{1}{{{{40}}}} = 125100\dfrac{{125}}{{100}}
Since the concentration is equal in both the cases, so the final solution will be neutral.
Thus, the correct answer is option A.

Note
The molar equivalent units represent the amounts in milligrams of a solute equal to 11000\dfrac{1}{{1000}} of the gram equivalent weight taking into consideration the valence of the ions present in the solution. While the equivalent weight of a solute = formula weightvalency\dfrac{{{\text{formula weight}}}}{{{{valency}}}} .