Consider titration of NaOH solution versus 1.25M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5mL (ii) 4.5mL (iii) 4.4mL (iv) 4.4mL (v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ______M. (Rounded-off to the nearest integer)

Answer

To find the molarity of the NaOH solution used in the titration with oxalic acid, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) can be represented as follows:

\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the normality of oxalic acid.
Oxalic acid is a diprotic acid, meaning it can donate 2 protons (H⁺ ions). Therefore, the equivalent factor (n-factor) for oxalic acid is 2.

Given:
- Volume of oxalic acid (V1) = 10.0 mL = 0.010 L
- Molarity of oxalic acid (M1) = 1.25 M

The normality (N1) of oxalic acid can be calculated as:
\[ N1 = M1 \times n = 1.25 \, \text{M} \times 2 = 2.5 \, \text{N} \]

### Step 3: Use the titration formula.
Using the formula \( N1 \times V1 = N2 \times V2 \), where:
- N1 = Normality of oxalic acid
- V1 = Volume of oxalic acid
- N2 = Normality of NaOH
- V2 = Volume of NaOH at the endpoint

### Step 4: Identify the volume of NaOH used.
From the given readings, the concordant volumes of NaOH are 4.4 mL (the last three readings). We will use this value for our calculation:
- V2 = 4.4 mL = 0.0044 L

### Step 5: Calculate the normality of NaOH.
Now we can rearrange the equation to find N2 (normality of NaOH):
\[ N2 = \frac{N1 \times V1}{V2} \]
Substituting the values:
\[ N2 = \frac{2.5 \, \text{N} \times 0.010 \, \text{L}}{0.0044 \, \text{L}} \]
\[ N2 = \frac{0.025}{0.0044} \approx 5.68 \, \text{N} \]

### Step 6: Calculate the molarity of NaOH.
Since NaOH is a strong base and has an n-factor of 1, the molarity (M2) is equal to the normality (N2):
\[ M2 = N2 = 5.68 \, \text{M} \]

### Step 7: Round off to the nearest integer.
Rounding 5.68 to the nearest integer gives:
\[ M2 \approx 6 \, \text{M} \]

Thus, the molarity of the NaOH solution is **6 M**.

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