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.

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.

Ans:

According to acid base titration, we know that equivalent amount of acid neutralized equivalent amount of base.

So, volume of HCl solution neutralized by NaOH

= (11.72 X 0.1506)/0.1034 ml

= 17.069 ml

Acid consumed by antacid = (32.36 - 17.069) ml

= 15.291 ml

NaOH solution add to this acid solution = 11.72 ml 0.1506M

1000 ml 0.001765M

Antacid added to this solution = equivalent amount of acid neutralized by antacid

= 15.291 ml 0.1034 M

1000 ml 0.001581 M

So, the total moles of hydroxide ion added to HCl solution

= No of moles antacid + no of moles NaOH

= 0.001581 M + 0.001765 M

= 0.003346 M

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