Two enterprising chemistry students decide to make an alternative to alka-seltze

Question

Two enterprising chemistry students decide to make an alternative to alka-seltzer tablets, replacing the asprin (acetylsalicylic acid) with vitamin C (ascorbic acid). The following table shoes one of the formulations they tried for a 5 g tablet.

Ingredient - Formula - Quantity (mg)

soduim bicarbonate - NaHCO3 - 1991

citric acid - H3C6H5O7 - 886

Vitamin C - HC6H7O6 - 370

inert binder - no formula given -1753

Total mass: 5000

When the tablet dissolves in water, it fizzes as the bicarbonate reacts with the citric acid and the vitamin C. What mass of the NaHCO3 reacts with the vitamin C?

Citric acid is triprotic, and reacts with base to form the citrate ion (C6H5O7^3-). Write the overall net ionic reaction for the acidification of NaHCO3 by citric acid, giving CO2 as the product. Citric acid should be written as H3C6H5O7 in the net ionic equation, as weak acids are written in the molecular form even though they slightly ionize.

What mass of the NaHCO3 reacts with the citric acid?

What is the mass of CO2 released in this reaction?

The remaining NaHCO3 is available to neutralize stomah acid. What mass of stomach HCl could be neutralized by one tablet?

I really don't understand this so all help is appreciated.

Explanation / Answer

mols of NaHCO3 = 1991mg*(1g/1000mg)*(1mol/84.007g i.e. Molar Mass) = 0.0237 mol NaHCO3;

mols of Citric Acid = 0.886g*(1mol/192.12g)=0.0046 mol Citric Acid;

mols of Vitamin C = 0.370g*(1mol/176.12g)=0.0021mol Vitamin C;

There is more moles of NaHCO3 than Citric Acid and Vitamin C combined so that means that Citric Acid and Vitamin C are the limiting reagents;

For question 1, we can assume that all the Vitamin C is used up in the reaction and that there is a 1:1 mol to mol ratio. So all we need to do is plug the moles of Vitamin C into the NaHCO3 molar mass and work backwards.

0.0021 mol (the total moles of Vitamin C) NaHCO3 * (84.007g NaHCO3 / 1mol)*(1000mg/1g) = 176.4mg

176.4 mg of NaHCO3 react with Vitamin C

For question 2, start by writing the TOTAL ionic equation and then cancel out the chemicals on both sides of the reaction to form the NET ionic equation. From the question, we know that Citric Acid's purpose is to donate protons (H+) and forms C6H5O7^3-. This tells us that HCO3 forms the CO2 and that we need three carbons for every Citric Acid Molecule.

TOTAL: Na^+ + 3 HCO3^- + C6H8O7 --> Na^+ + 3 CO2 + H2O + C6H5O7^3-

Na is on both sides so it will be removed in the NET ionic equation.

NET: 3HCO3^- + C6H8O7 --> 3 CO2 + 3 H2O + C6H5O7^3-

For question 3, we apply the same procedure we used for question 1 but with Citric Acid instead of Vitamin C. We also have to use the net ionic equation we just determined above for the mol-to-mol ratio between the two. There are 3 NaHCO3 ions for every Citric Acid molecule so we will need 3 times the moles of Citric Acid for NaHCO3.

0.0046mol Citric Acid * (3mol NaHCO3 / 1 mol Citric Acid) * (84.007g/1mol NaHCO3) * (1000mg/1g) = 1159mg NaHCO3

For question 4, we apply the same procedure as question 3 but with CO2. Once again we use the net ionic equation to see that there are 3 CO2 produced from 3 NaHCO3 so this is a 1-to-1 mol-to-mol ratio.

.0138 mol NaHCO3 * (1mol CO2/1mol NaHCO3)*(44.01g CO2/1mol)*(1000mg/1g) = 607mg CO2

For question 5, we need to think about acids and bases. Acids produce a high concentration of H+ in a solution so to neutralize, we need a base that can bind those H+. HCO3^- can bind 1 H+. So we need to set up a balanced equation.

TOTAL: HCO3^- + H+ + Cl^- --> H2CO3 + Cl^-

NET: HCO3^- + H+ --> H2CO3

There is a 1 to 1 mol-to-mol ratio.

1991mg total NaHCO3 - 176.4mg for Vit C - 1159mg for Citric Acid = 655.6 mg NaHCO3

0.6556g NaHCO3*(1mol NaHCO3/84.007g)*(1mol HCl/1mol NaHCO3)*(36.46g HCl/1mol)*(1000mg/1g) = 284.5mg HCl

The remaining HCO3^- can neutralize 284.5mg of HCl.

I used mg as the masses because it gave you mg initially.

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