Question

Laboratory analysis of a water sample reported the presence of following constituents \(\text{Na} = 14.8\text{ meq/l}\), \(\text{Ca} = 3.5\text{ meq/l}\), \(\text{Mg} = 4.5\text{ meq/l}\), \(\text{CO}_3 = 0.4\text{ meq/l}\), \(\text{HCO}_3 = 2.6\text{ meq/l}\). The sodium adsorption ratio of water is

• A. 7.4
• B. 11.2
• C. 9.9
• D. 8.5

Hint:

SAR formula: $\text{SAR} = \frac{\text{Na}}{\sqrt{(\text{Ca} + \text{Mg})/2}} = \frac{14.8}{\sqrt{8/2}} = \frac{14.8}{2} = 7.4$. (Low sodium hazard: $\text{SAR} < 10$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Sodium Adsorption Ratio (SAR) of irrigation water evaluates the sodium hazard and risk of soil sodification/clay dispersion.
Key Formula or Approach:
\[ \text{SAR} = \frac{[\text{Na}^+]}{\sqrt{\frac{[\text{Ca}^{2+}] + [\text{Mg}^{2+}]}{2}}} \]

Step 2: Detailed Explanation:
Given cation concentrations in meq/L:
- \([\text{Na}^+] = 14.8\text{ meq/L}\)
- \([\text{Ca}^{2+}] = 3.5\text{ meq/L}\)
- \([\text{Mg}^{2+}] = 4.5\text{ meq/L}\)

Step 1: Calculate the denominator:
\[ \frac{[\text{Ca}^{2+}] + [\text{Mg}^{2+}]}{2} = \frac{3.5 + 4.5}{2} = \frac{8.0}{2} = 4.0 \]
\[ \sqrt{4.0} = 2.0 \]
Compute SAR:
\[ \text{SAR} = \frac{14.8}{2.0} = 7.4 \]

Step 3: Final Answer:
Hence, the sodium adsorption ratio of the water sample is 7.4, corresponding to option (A).