How many kg / day of NaOH must be added to neutralize a waste stream generated by an industry producing 90,800 kg / day of sulfuric acid, if 0.1% of the sulfuric acid produced ends up in the wastewater? The wastewater flow is 750,000 L / day.

74.12 kg/day

Explanation:

First we write the complete neutralization reaction of sodium hydroxide and sulfuric acid. second, neglect wastewater flow rate since you must neutalize all the acid in the waste water. (acid and water make up the waste stream)

2NaOH + H2SO4 → Na2SO4 + 2H2O

Amount of sulfuric acid that enters wastewater = (0.1/100) * 90,800 kg/day

= 90.8kg/day

From the equation above; 98g of H2SO4 was neutralized by 80g of NaOH

90.8 kg/day in the waste stream will be neutralized by how many kg/day of NaOH?

Expressing the above statement in proportion;

98 g → 80 g

90.8 kg/day →?

= (90.8 kg/day * 80 g) / 98 g

0.1% of sulfuric produced in the waste stream will require 74.12 kg/day of NaOH

74.12kg/day

Explanation:

Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O

Mass of sulfuric acid produced per day = 90,800kg

Percentage of sulfuric acid in wastewater = 0.1%

Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 * 90,800 = 90.8kg

From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)

80kg of NaOH is required to neutralize 98kg of sulfuric acid

90.8kg of sulfuric acid would be neutralized by (90.8*80) / 98kg of NaOH = 74.14kg/day of NaOH