Chemistry question help is appreciated?

Question

a chemical plant that manufactures sulfuric acid produces 545L of wastewater per day that contain 1.25g/L of sulfuric acid.

a) what is the concentration of sulfuric acid is ppm?

b) what is the molarity of sulfuric acid

c) in order to meet the country's environmental regulation, the wastewater needs to be diluted to 2.04X10^-3M. what is the excess amount of water needed to dilute the wastewater?

d) the chemical plant manufactures sulfuric acid of a concentration of 33.5% (by weight) for making lead acid batteries used in cars. calculate the mole fraction of sulfuric acid

e) calcium hydroxide is added to sulfuric acid to precipitate the sulfate ions, what is the volume of 0.100M of calcium hydroxide needed to react completely with the sulfuric acid per day?

Answer 1

a chemical plant that manufactures sulfuric acid produces 545L of wastewater per day  

a) what is the concentration of sulfuric acid in ppm, that contains 1.25g/L of sulfuric acid?

this is 8X's weaker than 1% H2SO4, which the CRC handbook states has a density of 1.005 g/ml,... so we can use the density to be that of H2O itself, 1.00 g/ml:

the # of milligrams of stuff per liter of mix = ppm when density is 1g/ml

1.25 g/litre @ 1.00g/ml =>=> => 1.25 g / 1,000 grams

1.25 g / 1,000 grams => =>  => 1250 mg H2SO4 / 1,000,000 mg

solution

that's 1250 parts H2SO4 per million parts solution

your answer is 1250ppm

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b) what is the molarity of sulfuric acid, using molar mass:

1.25 g H2SO4 @ 98g/mol = 0.012755  moles in that litre of solution

your answer is 0.0128 Molar

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c) in order to meet the country's environmental regulation, the wastewater needs to be diluted to 2.04X10^-3M. what is the excess amount of water needed to dilute the wastewater?

using the dilution formula:

C1V1 = C2V2

(2.04e-3M)(V1) = ( 0.0128 M)(1 litre)

V1 = 6.27 litres of diluted H2SO4

6.27 litres diluted - 1 litre stock = 5.27Litres needed to be added

the answer that I don't think you want is: you need to add 5.27 litres of water to each litre of H2SO4 waste water

C1V1 = C2V2

(2.04e-3M)(V1) = ( 0.0128 M)(545 litres)

V1 = 3420 litres of diluted H2SO4

3420 litres diluted - 545 litres stock = 2874.6 litres needs to be added per day

the answer that I think you do want is: 2870 litres needs to be added per day (3sigfigs)

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d) the chemical plant manufactures sulfuric acid of a concentration of 33.5% (by weight) for making lead acid batteries used in cars. calculate the mole fraction of sulfuric acid

33.5 % => => 33.5 g H2SO4 @ 98.0g/mol = 0.342 moles H2SO4

66.5% => => 66.5 g H2O @ 18.0 g/mole = 3.694 moles H2O

mole fraction H2SO4: 0.342 moles H2SO4 / 4.036 total = 0.0847 H2SO4

mole fraction H2O: 3.694 moles H2O / 4.036 total = 0.915 H2O

your mole fraction answers are: 0.0847 H2SO4, &   0.915 H2O

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e) calcium hydroxide is added to sulfuric acid to precipitate the sulfate ions, what is the volume of 0.100M of calcium hydroxide needed to react completely with the sulfuric acid per day?

we can only calculate the neutralization of the H2SO4 in the waste water per day, because we don't know how much H2SO4 they produce for sale as battery acid:

0.012755  moles H2SO4 are in each litre of waste solution, :

545 litres  waste @ 0.012755  mol / litre = 6.951 moles of H2SO4 per day

since 1 Ca(OH)2  & 1 H2SO4 --> 1 CaSO4  & 2 H2O , (1:1ratio):

6.951 moles of H2SO4 per day = 6.951 moles of Ca(OH)2 per day

find the volume of 0.100M of calcium hydroxide needed to react completely with the sulfuric acid per day:

6.951 mol Ca(OH)2 @ 0.100 mol / litre = 69.5 litres Ca(OH)2

your answer is : you need to add 69.5 litres of ).100Molar Ca(OH)2 per day