The concentration of an oxalic acid solution is `x` mol `litre^(-1)` . 40mL of this solution reacts with 16 mL of 0.05 M acidified `KMnO_(4)`. What is the pH `x` M oxalic acid solution ? (Assume that oxalic acid dissociates completely.)

Options

A. (a) 1.3
B. (b) 1.699
C. (c) 1
D. (d) 2

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between oxalic acid (H₂C₂O₄) and acidified KMnO₄ can be represented as follows:

\[ \text{KMnO}_4 + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{CO}_2 + \text{H}^+ + \text{Mn}^{2+} + \text{H}_2\text{O} \]

The balanced equation is:

\[ 2 \text{KMnO}_4 + 5 \text{H}_2\text{C}_2\text{O}_4 \rightarrow 10 \text{CO}_2 + 8 \text{H}^+ + 2 \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 2: Determine the n-factor for KMnO₄ and H₂C₂O₄
- For KMnO₄, the oxidation state of Mn changes from +7 to +2, which means it gains 5 electrons. Thus, the n-factor for KMnO₄ is 5.
- For H₂C₂O₄, the oxidation state of carbon changes from +3 to +4, which means it loses 1 electron per carbon atom. Since there are 2 carbon atoms, the n-factor for H₂C₂O₄ is 2.

### Step 3: Calculate the milliequivalents of both reactants
Using the formula for milliequivalents:

\[ \text{Milliequivalents} = \text{Concentration (M)} \times \text{Volume (L)} \times \text{n-factor} \]

For KMnO₄:
\[ \text{Milliequivalents of KMnO}_4 = 0.05 \, \text{mol/L} \times 0.016 \, \text{L} \times 5 = 0.004 \, \text{equivalents} \]

For H₂C₂O₄:
\[ \text{Milliequivalents of H}_2\text{C}_2\text{O}_4 = x \, \text{mol/L} \times 0.040 \, \text{L} \times 2 \]

### Step 4: Set the milliequivalents equal to each other
Since the milliequivalents of both reactants are equal:

\[ 0.004 = x \times 0.040 \times 2 \]

### Step 5: Solve for x
Rearranging the equation to solve for x:

\[ 0.004 = 0.08x \]

\[ x = \frac{0.004}{0.08} = 0.05 \, \text{mol/L} \]

### Step 6: Calculate the concentration of H⁺ ions
Since oxalic acid dissociates completely, the concentration of H⁺ ions will be:

\[ [\text{H}^+] = 2 \times [\text{H}_2\text{C}_2\text{O}_4] = 2 \times 0.05 = 0.1 \, \text{mol/L} \]

### Step 7: Calculate the pH
The pH is calculated using the formula:

\[ \text{pH} = -\log[\text{H}^+] \]

Substituting the concentration of H⁺ ions:

\[ \text{pH} = -\log(0.1) = -\log(10^{-1}) = 1 \]

### Final Answer
The pH of the x M oxalic acid solution is **1**.

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