Question

125 ml of 63% (w/v) ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ solution is made to react with 125 mL of a 40%(w/v) NaOH solution. The resulting solution is:

• A. Neutral
• B. Acidic
• C. Strongly acidic
• D. Alkaline

Answer 1

The correct option is A

Neutral

Normality: The normality of a solution is defined as the number of equivalents of solute present in one liter (1000 ml) solution.

$\mathrm{Normality}\left(\mathrm{N}\right)=\frac{\mathrm{Number}\mathrm{of}\mathrm{Equivalent}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{litre}}$

Given 125 mL of 63% (w/v) ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ solution.

Given Weight of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ = 63 gm

The molecular weight of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$=126 g/mol.

The equivalent weight of ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ = $\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Basicity}}=\frac{126}{2}=63$

$\mathrm{Normality}\left(\mathrm{N}\right)\mathrm{of}{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4.2{\mathrm{H}}_2\mathrm{O}=\frac{63}{63}\times \frac{1000}{125}=8\mathrm{N}$

Given 125 mL of 40% (w/v) NaOH solution.

Weight of NaOH= 40 gm

The molecular weight of NaOH =40 g/mol.

Equivalent weight of NaOH = 40 /1 = 40

$\mathrm{Normality}\left(\mathrm{N}\right)\mathrm{of}\mathrm{NaOH}=\frac{40}{40}\times \frac{1000}{125}=8\mathrm{N}$

Since the normality of acid and base is equal in solution then the resultant solution is Neutral in nature.

Conclusion Statement: Hence, option(A) is correct option.