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In an industrial spill, 5, 000 L of water contaminated with nitric acid (a monop

ID: 919288 • Letter: I

Question

In an industrial spill, 5, 000 L of water contaminated with nitric acid (a monoprotic strong acid) is captured in a containment basin designed to minimize environmental damage. The contaminated water must be neutralized with CaCO_3 before it can be discharged: 2 HNO_3(aq) + CaCO_3(j) rightarrow Ca(NO_3)_2(aq) + CO_2(g) + H_2O(l) If the acidic water is found to have a pH of 1.7, what mass of CaCO_3 must be added to neutralize the water? A saturated (i.e., maximum dissolved solute possible") aqueous solution of ammonium sulfate is prepared by adding 150.35 g of the compound to enough water to form 250.0 mL of solution 19 degree C. After stirring for one hour, the solution is filtered and 42.65 g of undissolved dried compound is recovered. Calculate both the molar (M) and weight percent.(wt %) concentration of the solution. The density of the solution is 1.27 g/mL.

Explanation / Answer

Q:1: Since HNO3 is strong acid, it is completely dissociated into H+ and NO3-.

Hence [H+] = [NO3-] = [HNO3]

Given the volume of the contaminated water, V = 5000 L

pH of the contaminated water, pH = 1.7

=> - log[H+] = 1.7

=> [H+] = 10-1.7 = 0.019953 M

Hence [H+] = [NHO3] = 0.019953 M

Hence moles of HNO3 in the contaminated water = Concentration of HNO3 x V

= 0.019953 M x 5000 L = 99.77 mol HNO3

The balanced chemical reaction for the reaction of HNO3 with CaCO3 is

2 HNO3(aq) + CaCO3(s) --- > Ca(NO3)2 + CO2(g) + H2O(l)

2 mol ----------- 1 mol ------------- 1 mol --------1 mol -----1 mol

In the above balanced reaction, 2 mol of HNO3 reacts with 1 mol of CaCO3.

Hence 99.77 mol HNO3 that will react with the moles of CaCO3

= (1 mol CaCO3 / 2 mol HNO3) x 99.77 mol HNO3

= 49.88 mol CaCO3

Molecular mass of CaCO3 = 100.0 g/mol

Hence mass of CaCO3 required = 49.88 mol CaCO3 x (100.0 g / mol) = 4988 g CaCO3 (answer)

= 5000 g CaCO3 (upto 1 SF) (answer)

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