Question

# A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl....

A student dissolves 0.326 g of a powdered antacid in 32.36 mL of 0.1034 M HCl. The student boils the mixture and then allows it to cool. Lastly, the student adds phenolphthalein indicator to the mixture.

1) Calculate the total number of moles of H+ added to the antacid.

2)Suppose 11.72 mL of 0.1506 M NaOH is required to turn the solution from colorless to pale pink. Calculate the total moles of OH- added.

3) Calculate the difference between total moles of H+ added and total moles of OH- added.

4) How many moles of HCl reacted with the antacid? How many equivalents of antacid are present in the sample?

5) Find the number of antacid equivalents present per gram of antacid. Note the number of moles of HCl that react with antacid equals the number of antacid equivalents present.

6) Given an antacid costs \$5.99 per 100 tablet bottle and the average mass of a tablet is 650 mg, calculate the cost per equivalent (in \$/eq) of this antacid.

Ans

1)

Number of moles of H+ = molarity of HCl x volume (L)

= 0.1034 x 0.03236

= 0.003346 mol

2)

The total moles of OH- = moalrity x volume (L)

= 0.1506 x 0.01172

= 0.001765 mol

3)

Difference = 0.003346 - 0.001765

= 0.001581 moles

4)

Since 0.001765 moles of NaOH reacted with same number of moles of HCl , so the moles of HCl reacted with antacid will be 0.001581 moles.

Each HCl mole react with one equivalent of antacid. So number of equivalents of antacid in sample will be 0.001581 eq.

5)

Number of antacids per gram of antacid = 0.001581 / 0.326

= 0.00485 eq antacid per gram

6)

0.001581 eq is present in 0.326 g or 326 mg of antacid

so eq present in 650 mg tablet : ( 650 x 0.001581) / 326

= 0.003152 eq per tablet

Eq in 1 bottle : 0.003152 x 100

= 0.3152 eq/bottle

So the cost per equivalent : 5.99 / 0.3152

= 19 \$/eq

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