Question 10.1: DETERMINATION OF SODIUM ADSORPTION RATIO Determine the SAR f......
DETERMINATION OF SODIUM ADSORPTION RATIO
Determine the SAR for two different water supplies, A and B. Measured concentrations were
Supply A Ca^{2+}=4.6 mg/L Mg^{2+}=1.1 mg/L Na^{+}=144 mg/L TDS=255 mg/L
Supply B Ca^{2+}=108 mg/L Mg^{2+}=33 mg/L Na^{+}=63 mg/L TDS=1050 mg/L
1. Determine the Ca^{2+}, Mg^{2+}, and N^{2+} concentrations in meq/L (see discussion of equivalent weight in Chapter 1).
2. Compute the SAR of the two waters.
a. For Supply A
This is a very high SAR value, and the water would be generally unacceptable for irrigation. Note that the TDS value for water supply A is moderate (255 mg/L) and that Supply A water is generally excellent for domestic use.
b. For Supply B
Water supply B has a low SAR value and a TDS value (1050 mg/L,
equivalent to about 1750 \mu S/cm, see Chapter 3, Section 3.12.3) that is beneficial for agriculture. It would be an excellent source of irrigation water. The TDS value is high for potable water and water supply B is an inferior source of domestic water compared with Supply A. Supply B will require more treatment to produce acceptable finished drinking water.
| Concentration | ||||||
| Constituent | Equivalent Weight (g/eq or mg/meq) |
Water Supply A | Water Supply B | |||
| (mg/L) | (mg/L) | (mg/L) | (mg/L) | |||
| Ca^{2+} | 20.0 | 4.6 | 0.23 | 108 | 5.40 | |
| Mg^{2+} | 12.15 | 1.1 | 0.090 | 33 | 2.72 | |
| Na^{+} | 23.0 | 255 | 11.1 | 63 | 2.74 | |