Question

Two enterprising chemistry students decide to develop an alternative to Alka-Seltzer tablets, replacing the aspirin (acetylsalicylic acid) with vitamin c (ascorbic acid). The following table shows one of the formulations they tried for a 5 g tablet.

ingredient	formula	Quantity (mg)
Sodium bicarbonate	NaHCO3	1991
citric acid	H3C6H5O7	886
vitamin c	HC6H7O6	370
inert binder		1753
TOTAL MASS:		5000

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

Citric acid is triprotic, and reacts with the base to form the citrate ino (C₆H₅O₇^3-). Write the overall ionic reaction for the acidification of the NaHCO₃ by the 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 neautralize stomach acid. What mass of stomach HCL could be neutralized by one tablet?

Answer 1

What mass of NaHCO₃ reacts with the vitamin C?

Since vitamin c is monoprotic acid and so mole ratio between NaHCO3 and vitamin C is 1 :1

Moles of vitamin c = 370 E -3 g / molar mass of vitamin c = 370 E-3 g / 176.12 g per mol = 0.0021 mol

Moles of NaHCO3 = 0.0021 mol NaHCO3 = 0.0021 mol x 80.007 g per mol = 0.168 g = 168 mg

Citric acid is triprotic, and reacts with the base to form the citrate ino (C₆H₅O₇^3-). Write the overall ionic reaction for the acidification of the NaHCO₃ by the citric acid, giving CO₂ as the product.

3NaHCO3(s) + H3C6H5O7(aq)   -- > Na3C6H5O7(aq) + 3CO2(g) + 3H2O(l)

Ionic equation :

3NaHCO3(s) + H3C6H5O7(aq)   -- > 3 Na+ (aq) + C6H5O7^3- (aq) + 3CO2(g) + 3H2O(l)

What mass of the NaHCO₃ reacts with the citric acid?

Calculation of mass :

moles of NaHCO3 = (mass of citric acid / molar mass ) x ( 3 mol NaHCO3 / 1 mol citric acid )

= (0.886 g / 192.124 g per mol )x ( 3 mol NaHCO3 / 1 mol Citric acid )

=0.0138 mol NaHCO3

Mass of NaHCO3 = 0.0138 mol x 80.007 g /mol = 1.11 g

What is the mass of CO₂ released in this reaction

Moles of CO2 = Moles of NaHCO3 x 3 mol CO2 / 3 mol NaHCO3

= 0.0138 mol CO2

Mass of CO2 = 0.0138 mol x 44.009 g per mol= 0.61 g

THe remaining NaHCO₃ is available to neautralize stomach acid. What mass of stomach HCL could be neutralized by one tablet?

Reacted mass of NaHCO3 = 1.11 g

Excess mass of NaHCO3 = 1.991 g - 1.11 g = 0.88 g

Calculation of moles of NaHCO3 = 0.88 g / 84.007 g per mol= 0.0105 mol

Mole ration of reaction of sodium carbonate with HCl is 1 : 1 since HCl is monoprotic acid

Therefore moles of HCl = moles of sodium bicarbonate = 0.0105

Mass of HCl = 0.0105 mol x 36.46 g /mol

= 0.384 g

Mass of HCl = 0.384 g = 384 mg mass is neutralized by NaHCO3