Answer:
The HCl is very strong since its pH is equal to 0.17.
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:
[tex] n_{CaCO_{3}} = \frac{m}{M} [/tex]
Where:
m: is the mass = 0.750 g
M: is the molar mass = 100.0869 g/mol
[tex] n_{CaCO_{3}} = \frac{0.750 g}{100.0869 g/mol} = 7.49 \cdot 10^{-3} moles [/tex]
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:
[tex] n_{HCl} = 2*7.49 \cdot 10^{-3} moles = 0.015 moles [/tex]
Now, with the number of moles of HCl we can find its concentration:
[tex] C = \frac{n}{V} = \frac{0.015 moles}{22.25 \cdot 10^{-3} L} = 0.67 M [/tex]
Finally, the pH of the acid is:
[tex] pH = -log([H^{+}]) = -log(0.67) = 0.17 [/tex]
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!
