Question

Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained. (i) 4.5 mL, (ii) 4.5 mL, (iii) 4.4 mL, (iv) 4.4 mL, (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)

Hint:

In titration calculations, always start by writing the balanced chemical equation to determine the stoichiometric ratio (or n-factors). The formula $n_a M_a V_a = n_b M_b V_b$ is essential. If your calculated answer doesn't match the key, double-check the n-factors and then look for potential typos in the given data.

Answer 1

Correct Answer: 6

Step 1: Write the balanced chemical equation Oxalic acid reacts with sodium hydroxide as: \[ \mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O} \] From the equation:
1 mole of oxalic acid reacts with 2 moles of NaOH
n-factor of oxalic acid, $n_a = 2$
n-factor of NaOH, $n_b = 1$
Step 2: Determine the concordant burette reading The burette readings obtained are: \[ 4.5,\; 4.5,\; 4.4,\; 4.4,\; 4.4\ \text{mL} \] The concordant (most consistent) reading is: \[ V_b = 4.4\ \text{mL} \] Step 3: Note the given data \[ M_a = 1.25\ \text{M}, \quad V_a = 10.0\ \text{mL}, \quad V_b = 4.4\ \text{mL} \] Step 4: Apply the titration formula \[ n_a M_a V_a = n_b M_b V_b \] Substituting values: \[ 2 \times 1.25 \times 10.0 = 1 \times M_b \times 4.4 \] Step 5: Calculate the molarity of NaOH \[ 25 = 4.4 M_b \] \[ M_b = \frac{25}{4.4} \approx 5.68\ \text{M} \] Step 6: Rounding off The question asks for the answer rounded to the nearest integer: \[ M_b \approx \boxed{5} \] Final Answer: \[ \boxed{5\ \text{M}} \]