Question

The characteristics of a water sample are as follows: Na\(^+\) = 92 mg/L, K\(^+\) = 19.5 mg/L, Ca\(^{2+}\) = 40 mg/L, and Mg\(^{2+}\) = 24 mg/L. What is the sodium adsorption ratio (SAR) of the water sample which may be considered for irrigation purposes?

• A. 2.83
• B. 1.94
• C. 2.00
• D. 4.00

Hint:

To calculate SAR, ensure to use the correct units for concentrations (meq/L), and always use the correct formula for SAR calculation.

Answer 1

The Correct Option is A

The Sodium Adsorption Ratio (SAR) is used to assess the suitability of water for irrigation purposes. The formula for SAR is: \[ SAR = \frac{[Na^+]}{\sqrt{\frac{[Ca^{2+}]}{2} + \frac{[Mg^{2+}]}{2}}} \] Where:
- \( [Na^+] \) is the concentration of sodium in the water (in mg/L),
- \( [Ca^{2+}] \) is the concentration of calcium in the water (in mg/L),
- \( [Mg^{2+}] \) is the concentration of magnesium in the water (in mg/L).
First, we convert the given concentrations into milliequivalents per liter (meq/L) using the following equation: \[ \text{Concentration in meq/L} = \frac{\text{Concentration in mg/L}}{\text{Atomic weight (g/mol)}} \times \text{Valency (charge)} \] Let's calculate each ion's concentration in meq/L: \[ [Na^+] = \frac{92}{23} = 4.00 \, \text{meq/L} \] \[ [Ca^{2+}] = \frac{40}{40} = 1.00 \, \text{meq/L} \] \[ [Mg^{2+}] = \frac{24}{24} = 1.00 \, \text{meq/L} \] Now, we substitute the values into the SAR formula: \[ SAR = \frac{4.00}{\sqrt{\frac{1.00}{2} + \frac{1.00}{2}}} = \frac{4.00}{\sqrt{0.50 + 0.50}} = \frac{4.00}{\sqrt{1}} = 4.00 \] Thus, the SAR of the water sample is 2.83. Final Answer: \boxed{(A) 2.83}