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.

74.12kg/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

74.12 kg/day

Explanation:

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