On the box of Arm and Hammer baking soda it says the contents are sodium bicarbo

Question

On the box of Arm and Hammer baking soda it says the contents are sodium bicarbonate U.S.P, (United States Pharmacopeia). Basically, this means the contents are pure sodium bicarbonate. The box also says that baking soda can be used as an antacid and says that one dose is 1/2 level teaspoon in 4 fl. Oz. of water.

1. What is the neutralizing power of one dose of baking soda? One dose is 1/2 teaspoon of baking soda dissolved in 4 fl. Oz. of water. Use 0.71g/mL as the density of granular baking soda. 1 teaspoon=4.93mL.

2. Which is cheaper per mole of hydrochloric acid neutralized, a generic antacid costs $8.99 for a bottle of 500 tablets each of which contains 1000mg of CaCO3 or a 1 pound box of baking soda that costs $0.69 per box?

Explanation / Answer

1) First, the respective calculations are made to determine the concentrations of NaHCO3 and HCl.

1/2 teaspoon * 4.93 ml * (0.71 gr / ml) = 1.75 gr * (1 mol / 84 gr) = 0.021 mol

[NaHCO3] = 0.021 mol / 0.12 L = 0.175 M

assuming that in the stomach of the human being the pH = 2 we have:

[HCl] = Antilog (-2) = 0.01 M

For the reaction:

NaHCO3 + HCl = NaCl + H2O + CO2

Ka = 4.68 * 10 ^ -11

where:

[NaHCO3] = 0.175 - x

[HCl] = 0.01 - x

[NaCl] = [H2O] = [CO2] = x

Ka = x ^ 3 / (0.175-x) * (0.01-x)

clearing x we ??have: the concentration of neutralized HCl

x = 4.34 * 10 ^ -5 M

To determine the neutralized / Initial HCl ratio:

4.34 * 10 ^ -5 / 0.01 = 4.34 * 10 ^ -3 neutralized mol / initial mol HCl

this represents the neutralisation power of NaHCO3

2) Calculating the cost of each neutralized mole,

4.34 * 10 ^ -5 mol neutralized / 1.75 gr NaHCO3 = 2.48 * 10 ^ -5 mol / gr NaHCO3

Cost: (1 gr NaHCO3 / 2.48 * 10 ^ -5 mol HCl) * (1 pound / 453.6 g) * (0.69 $ / 1 pound) = 61.34 $ / mol neutralized HCl

For CaCO3:

The reaction that is fulfilled is:

CaCO3 + 2HCl = H2O + CO2 + CaCl2

($ 8.99 / 500 tablets) * (1 tablet / 1000 mg) * (1000 mg / 1 g) * (100 g / 1 mol CaCO3) * (1 mol CaCO3 / 2 mol HCl) = 0.9 $ / mol Neutralized HCl

Comparing the costs of neutralized HCl mole, we can conclude that it is more economical to buy CaCO3.

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