Subtopic: Leaching requirement and salinity control
Question Statement:
A sodic soil has a Cation Exchange Capacity of $CEC = 25.0\text{ meq/100 g}$ of soil. The initial Exchangeable Sodium Percentage is $ESP_i = 32.0\%$ and the target is to reduce it to $ESP_f = 10.0\%$. The depth of soil to be reclaimed is $D = 30.0\text{ cm}$ ($0.30\text{ m}$) with a dry bulk density of $\rho_b = 1.50\text{ g/cm}^3$ ($1500.0\text{ kg/m}^3$). Taking the equivalent weight of agricultural gypsum ($\text{CaSO}_4 \cdot 2\text{H}_2\text{O}$) as $86.08\text{ g/equivalent}$, the theoretical Gypsum Requirement ($GR$) in metric tons per hectare ($\text{t/ha}$) is ________ (round off to one decimal place).
Answer Key:
21.3
Step-by-Step Solution & Explanation
**Method 1: Equivalent Mass Exchange Formulation**
1. Mass of soil in $1.0\text{ ha}$ ($10,000\text{ m}^2$) to depth $D = 0.30\text{ m}$:
$M_{\text{soil}} = 10,000\text{ m}^2 \times 0.30\text{ m} \times 1500.0\text{ kg/m}^3 = 4,500,000\text{ kg} = 4.50 \times 10^9\text{ g}$
2. Exchangeable sodium to be replaced per $100\text{ g}$ of soil:
$\Delta ESP = ESP_i - ESP_f = 32.0\% - 10.0\% = 22.0\% = 0.220$
$\Delta \text{Na} = 0.220 \times 25.0\text{ meq/100 g} = 5.50\text{ meq/100 g} = 0.0550\text{ meq/g}$
3. Total milliequivalents of sodium to replace in $1.0\text{ ha}$:
$\text{Total meq} = 0.0550\text{ meq/g} \times (4.50 \times 10^9\text{ g}) = 2.475 \times 10^8\text{ meq} = 2.475 \times 10^5\text{ equivalents}$
4. Mass of gypsum required:
$M_{\text{gypsum}} = (2.475 \times 10^5\text{ eq}) \times 86.08\text{ g/eq} = 2.13048 \times 10^7\text{ g}$
$M_{\text{gypsum}} = 21,304.8\text{ kg} \approx 21.3\text{ metric tons/ha}$
**Method 2: Standard Agronomic Multiplier Shortcut**
$GR = \Delta \text{Na (meq/100g)} \times 3.874 = 5.50 \times 3.874 = 21.307\text{ t/ha} \approx 21.3\text{ t/ha}$