There are 109.16 grains of sulfuric anhydride (SO3) per gallon of rain water in

Question

There are 109.16 grains of sulfuric anhydride (SO3) per gallon of rain water in Glasgow rain. Each mole of SO3 becomes a mole of sulfuric acid H2SO4. 1. Compute the molar concentration of SO3. Note that one British grain=.0648 grams, and one British gallon=4.546 liters.
2. Assuming that each mile of SO3 becomes one mole of H2SO4, and that the sulfuric acid dissociates completely, estimate the pH of Glasgow rain.
3. The acid in the water was actually sulfurous acid (H2SO3) rather than sulfuric acid. What would the pH be if the [SO3] concentration computed above (#1) was equivalent to H2SO3? The acid dissociation constant for H2SO3<->H + HSO3 is Ka=1.54x10^-2. (Hint: considering only the first dissociation constant is sufficient for this problem; however, assuming that equilibrium [H2SO3] is the same as initial [H2SO3] is not permissible.)
4. Calculate the mass of washing soda (sodium carbonate) required to neutralize one British gallon of rainwater. There are 109.16 grains of sulfuric anhydride (SO3) per gallon of rain water in Glasgow rain. Each mole of SO3 becomes a mole of sulfuric acid H2SO4. 1. Compute the molar concentration of SO3. Note that one British grain=.0648 grams, and one British gallon=4.546 liters.
2. Assuming that each mile of SO3 becomes one mole of H2SO4, and that the sulfuric acid dissociates completely, estimate the pH of Glasgow rain.
3. The acid in the water was actually sulfurous acid (H2SO3) rather than sulfuric acid. What would the pH be if the [SO3] concentration computed above (#1) was equivalent to H2SO3? The acid dissociation constant for H2SO3<->H + HSO3 is Ka=1.54x10^-2. (Hint: considering only the first dissociation constant is sufficient for this problem; however, assuming that equilibrium [H2SO3] is the same as initial [H2SO3] is not permissible.)
4. Calculate the mass of washing soda (sodium carbonate) required to neutralize one British gallon of rainwater. 1. Compute the molar concentration of SO3. Note that one British grain=.0648 grams, and one British gallon=4.546 liters.
2. Assuming that each mile of SO3 becomes one mole of H2SO4, and that the sulfuric acid dissociates completely, estimate the pH of Glasgow rain.
3. The acid in the water was actually sulfurous acid (H2SO3) rather than sulfuric acid. What would the pH be if the [SO3] concentration computed above (#1) was equivalent to H2SO3? The acid dissociation constant for H2SO3<->H + HSO3 is Ka=1.54x10^-2. (Hint: considering only the first dissociation constant is sufficient for this problem; however, assuming that equilibrium [H2SO3] is the same as initial [H2SO3] is not permissible.)
4. Calculate the mass of washing soda (sodium carbonate) required to neutralize one British gallon of rainwater.

Explanation / Answer

1. mass of SO3 = (109.16)(0.0648) g = 7.07 g

molar mass of SO3 = 80 g/mol

no. of moles of SO3 = (7.07 g)/ (80 g/mol) = 0.0884 mol

molar concentration = (no. of moles of SO3)/volume of rain water

= 0.0884 mol/ 4.546 L = 0.0194 M

2. H2SO4 -----> 2 H+ + SO42-

Assuming each mole of SO3 becomes one mole of H2SO4 and that sulfuric acid dissociates completely, we get

[SO3] = [H2SO4 ]

and, 2 [H2SO4 ] = [H+]    (since each acid molecule give two H+ ions, [H+] is twice that of[H2SO4 ])

Therefore, [H+] = 2 [SO3] = 2(0.0194) = 0.0388 M

pH = -log[H+] = -log(0.0388) = 1.41

3.     initial concentration of H2SO3 = [SO3] = 0.0194 M

                     H2SO3 <---> H+ + HSO3-                     Ka=1.54x10-2

Assuming second dissociation is very amall, if x moles of H2SO3 has dissociated, then at equilibrium:

[H2SO3] = 0.0194 - x

[ H+] = [HSO3-] = x

Therefore, Ka = [ H+]. [HSO3-] / [H2SO3]

or, x2 /(0.0194 - x) = 1.54 x 10-2

x2 = 0.0154(0.0194 - x) = 0.000299 - 0.0154 x

solving for x, we get:

x = 0.011 M = [H+]

pH = -log(0.011) = 1.96 M

4. one mole of sulfuric acid (or sulfurous acid) consumes one mole of sodium carbonate.

H2SO4 + Na2CO3 ---> Na2SO4 +   2 H2CO3

[H2SO4] = 0.0194 M

no. of moles of sulfuric acid (or sulfurous acid) in 1 gallon of rain water = (0.0194 mol/L)(4.546 L) = 0.0882 mol

no. of moles of sodium carbonate required = 0.0882 mol

mass of sodium carbonate required = (106 g/mol)(0.0882 mol) = 9.35 g

Hence, 9.35 g of washing soda is needed to completely neutralize one British gallon of rain water.

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