Question

A 200 mL sample of water has an initial pH = 9. In order to determine alkalinity, the sample was titrated using 0.02 N \( \text{H}_2\text{SO}_4 \) acid to an end point of pH = 4.5. In the titration, 25 mL of 0.02 N \( \text{H}_2\text{SO}_4 \) acid was required. What is the total alkalinity of the sample in ‘mg/L as NaHCO\(_3\)’? [Atomic weight (g/mol): Ca = 40, Na = 23, H = 1, C = 12, S = 32, and O = 16]

• A. 20
• B. 125
• C. 210
• D. 305

Hint:

To calculate alkalinity, use the titration volume and normality of acid, along with the equivalent weight of NaHCO\(_3\) for the conversion.

Answer 1

The Correct Option is C

The alkalinity of water can be calculated from the titration data using the equation: \[ \text{Alkalinity (mg/L as NaHCO}_3\text{)} = \frac{\text{N of acid} \times \text{Volume of acid (mL)} \times \text{Equivalent weight of NaHCO}_3}{\text{Volume of sample (mL)}} \] First, let's calculate the equivalent weight of NaHCO\(_3\). The equivalent weight of NaHCO\(_3\) is the molar mass of NaHCO\(_3\) divided by 1 (since NaHCO\(_3\) provides 1 mole of H\(^+\)): \[ \text{Equivalent weight of NaHCO}_3 = \frac{(23 + 1 + 12 + 3 \times 16)}{1} = 84 \, \text{g/mol}. \] Now, calculate the alkalinity: \[ \text{Alkalinity} = \frac{0.02 \, \text{N} \times 25 \, \text{mL} \times 84 \, \text{g/mol}}{200 \, \text{mL}}. \] First, calculate the numerator: \[ 0.02 \times 25 \times 84 = 42. \] Now, divide by the volume of the sample: \[ \frac{42}{200} = 0.21. \] Convert to mg/L as NaHCO\(_3\): \[ 0.21 \times 1000 = 210 \, \text{mg/L}. \] Final Answer: \[ \boxed{210 \, \text{mg/L}}. \]