Q11Engineering Chemistry

Question

Calculate the amount of lime and soda required to soften 10,000L of a water sample containing the following impurities :

$C a (H C O_{3})_{2} = 16.2 mg/L$
$M g (H C O_{3})_{2} = 14.6 mg/L$
$C a S O_{4} = 13.6 mg/L$
$M g S O_{4} = 12.0 mg/L$
$M g C l_{2} = 9.5 mg/L$

Answer 1

The Lime-Soda process is a vital industrial chemical method for softening hard water. It involves adding calculated amounts of slaked lime ( $C a (O H)_{2}$ ) and soda ash ( $N a_{2} C O_{3}$ ) to precipitate dissolved calcium and magnesium salts as highly insoluble calcium carbonate ( $C a C O_{3}$ ) and magnesium hydroxide ( $M g (O H)_{2}$ ). Before calculating the chemical requirements, every dissolved impurity must be mathematically converted into its equivalent amount of $C a C O_{3}$ to provide a uniform basis for calculation, because $C a C O_{3}$ has a convenient molecular weight of $100$ and equivalent weight of $50$ .

Step 1: Convert Impurities to CaCO3 Equivalents

The formula used is: $C a C O_{3} equivalent = \frac{Mass of Impurity \times 100}{Molecular Weight of Impurity}$

Calcium Bicarbonate, $C a (H C O_{3})_{2}$ : $16.2 mg/L$ . Mol. wt = $162$ . $C a C O_{3} eq = 16.2 \times \frac{100}{162} = 10 mg/L$ . (Requires 1 mole of Lime per mole)
Magnesium Bicarbonate, $M g (H C O_{3})_{2}$ : $14.6 mg/L$ . Mol. wt = $146$ . $C a C O_{3} eq = 14.6 \times \frac{100}{146} = 10 mg/L$ . (Requires 2 moles of Lime per mole)
Calcium Sulfate, $C a S O_{4}$ : $13.6 mg/L$ . Mol. wt = $136$ . $C a C O_{3} eq = 13.6 \times \frac{100}{136} = 10 mg/L$ . (Requires 1 mole of Soda per mole)
Magnesium Sulfate, $M g S O_{4}$ : $12.0 mg/L$ . Mol. wt = $120$ . $C a C O_{3} eq = 12.0 \times \frac{100}{120} = 10 mg/L$ . (Requires 1 mole of Lime + 1 mole of Soda)
Magnesium Chloride, $M g C l_{2}$ : $9.5 mg/L$ . Mol. wt = $95$ . $C a C O_{3} eq = 9.5 \times \frac{100}{95} = 10 mg/L$ . (Requires 1 mole of Lime + 1 mole of Soda)

Step 2: Calculate Lime Requirement

Lime is required to deal with temporary hardness (bicarbonates) and any magnesium salts. Magnesium salts always require lime to precipitate as $M g (O H)_{2}$ , regardless of whether they are temporary or permanent. The factor $\frac{74}{100}$ relates the molecular weight of lime ( $74$ ) to $C a C O_{3}$ ( $100$ ).

Lime Req. (mg/L) = \frac{74}{100} \times [C a (H C O_{3})_{2} + 2 \times M g (H C O_{3})_{2} + M g S O_{4} + M g C l_{2}]

Lime Req. = \frac{74}{100} \times [10 + (2 \times 10) + 10 + 10] = \frac{74}{100} \times 50 = 37 mg/L

For a total volume of $10, 000 Liters$ : Total Lime = $37 mg/L \times 10, 000 L = 370, 000 mg = 0.370 kg$ .

Step 3: Calculate Soda Requirement

Soda ash is strictly required to remove permanent calcium hardness. This includes the original $C a S O_{4}$ , as well as the 'new' permanent calcium hardness generated when lime reacted with the permanent magnesium salts ( $M g S O_{4}$ and $M g C l_{2}$ ). The factor $\frac{106}{100}$ relates the molecular weight of soda ( $106$ ) to $C a C O_{3}$ ( $100$ ).

Soda Req. (mg/L) = \frac{106}{100} \times [C a S O_{4} + M g S O_{4} + M g C l_{2}]

Soda Req. = \frac{106}{100} \times [10 + 10 + 10] = 1.06 \times 30 = 31.8 mg/L

For a total volume of $10, 000 Liters$ : Total Soda = $31.8 mg/L \times 10, 000 L = 318, 000 mg = 0.318 kg$ .

Conclusion: To fully soften the batch of $10, 000 liters$ , the plant operator must precisely add $0.37 kg$ of pure Lime and $0.318 kg$ of pure Soda Ash.

Q11Engineering Chemistry

Question

Answer

Step 1: Convert Impurities to CaCO3 Equivalents

Step 2: Calculate Lime Requirement

Step 3: Calculate Soda Requirement