Question

an extra strength antacid tablet contains 750 mg of active ingredient, caco3. if it takes 22.25 ml of hcl to neutralize the tablet, how strong is the acid

Answer 1

The HCl is very strong since its pH is equal to 0.17.

Step-by-step explanation:

The reaction between CaCO₃ and HCl is:

CaCO₃(s) + 2HCl(aq) ⇄ CaCl₂(aq) + CO₂(g) + H₂O(l) (1)

The number of moles of CaCO₃ is:

`n_{CaCO_(3)} = (m)/(M)`

Where:

m: is the mass = 0.750 g

M: is the molar mass = 100.0869 g/mol

`n_{CaCO_(3)} = (0.750 g)/(100.0869 g/mol) = 7.49 \cdot 10^(-3) moles`

From the reaction (1) we have that 1 mol of CaCO₃ reacts with 2 moles of HCl, so the number of moles of HCl is:

`n_(HCl) = 2*7.49 \cdot 10^(-3) moles = 0.015 moles`

Now, with the number of moles of HCl we can find its concentration:

`C = (n)/(V) = (0.015 moles)/(22.25 \cdot 10^(-3) L) = 0.67 M`

Finally, the pH of the acid is:

`pH = -log([H^(+)]) = -log(0.67) = 0.17`

The pH obtained is very low, so the HCl is very strong.

Therefore, the HCl is very strong since its pH is equal to 0.17.

I hope it helps you!