Question

3.15 g oxalic acid $[(COOH)_2.xH_2O]$ was dissolved to make 500 mL solution. On titration, 33.36 mL of this solution were neutralised by 50 mL $N/15$ NaOH. The value of $x$ will be

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

Answer: (B)

2

Answer 2

The reaction between oxalic acid $[(COOH)_2.xH_2O]$ and NaOH is a neutralization reaction. Oxalic acid is a dibasic acid, so its n-factor is 2.

The normality of the NaOH solution is given as $N/15$, which means $1/15$ N. The volume of NaOH solution used is 50 mL.

The milliequivalents (meq) of NaOH used in the titration are: meq of NaOH = Normality of NaOH $\times$ Volume of NaOH (in mL) meq of NaOH = $(1/15 \, \text{N}) \times (50 \, \text{mL}) = 50/15 \, \text{meq} = 10/3 \, \text{meq}$.

At the equivalence point of the titration, the milliequivalents of oxalic acid in the titrated volume are equal to the milliequivalents of NaOH used. The volume of oxalic acid solution titrated was 33.36 mL. meq of oxalic acid in 33.36 mL = $10/3 \, \text{meq}$.

The normality of the oxalic acid solution is: Normality of oxalic acid = $\frac{\text{meq of oxalic acid}}{\text{Volume of oxalic acid solution (in mL)}} = \frac{10/3 \, \text{meq}}{33.36 \, \text{mL}}$.

The total volume of the oxalic acid solution prepared was 500 mL. The total milliequivalents of oxalic acid in the 500 mL solution are: Total meq of oxalic acid = Normality of oxalic acid $\times$ Total volume of solution (in mL) Total meq of oxalic acid = $\left(\frac{10/3}{33.36}\right) \times 500 \, \text{meq}$.

The initial mass of hydrated oxalic acid $[(COOH)_2.xH_2O]$ dissolved to make the 500 mL solution was 3.15 g. The relationship between mass, milliequivalents, and equivalent weight is: Total meq = $\frac{\text{Mass (in g)}}{\text{Equivalent weight (in g/eq)}} \times 1000$ So, Equivalent weight = $\frac{\text{Mass (in g)} \times 1000}{\text{Total meq}}$ Equivalent weight of $[(COOH)_2.xH_2O]$ = $\frac{3.15 \, \text{g} \times 1000}{\left(\frac{10/3}{33.36}\right) \times 500 \, \text{meq}}$ Equivalent weight = $\frac{3.15 \times 1000 \times 33.36 \times 3}{10 \times 500} = \frac{3.15 \times 1000 \times 33.36 \times 3}{5000}$ Equivalent weight = $\frac{3.15 \times 33.36 \times 3}{5} = \frac{315.252}{5} = 63.0504 \, \text{g/eq}$.

The molar mass of anhydrous oxalic acid $(COOH)_2$ is approximately $2 \times (12 + 2 \times 16 + 1) = 2 \times 45 = 90$ g/mol. The molar mass of water $H_2O$ is approximately $2 \times 1 + 16 = 18$ g/mol. The molar mass of hydrated oxalic acid $[(COOH)_2.xH_2O]$ is approximately $(90 + 18x)$ g/mol. The equivalent weight of oxalic acid is its molar mass divided by its n-factor (which is 2). Equivalent weight = $\frac{\text{Molar mass}}{2} = \frac{90 + 18x}{2}$.

Equating the calculated equivalent weight from the titration data to the expression in terms of x: $\frac{90 + 18x}{2} = 63.0504$ $90 + 18x = 2 \times 63.0504 = 126.1008$ $18x = 126.1008 - 90 = 36.1008$ $x = \frac{36.1008}{18} \approx 2.0056$.

Since x must be an integer representing the number of water molecules of hydration, the value of x is approximately 2. The slight deviation from exactly 2 is likely due to experimental variations or rounding in the provided data. Oxalic acid commonly crystallizes as the dihydrate.

Assuming x = 2, the equivalent weight would be $(90 + 18 \times 2) / 2 = (90 + 36) / 2 = 126 / 2 = 63$. If the equivalent weight is exactly 63, the total meq in 500 mL is $(3.15 / 63) \times 1000 = 0.05 \times 1000 = 50$ meq. The normality of the solution is $50 \, \text{meq} / 500 \, \text{mL} = 0.1 \, \text{N}$. The meq in 33.36 mL is $0.1 \, \text{N} \times 33.36 \, \text{mL} = 3.336 \, \text{meq}$. The meq of NaOH used was $10/3 = 3.333... \, \text{meq}$. The values 3.336 and 3.333... are very close, confirming that x=2 is the correct value.