<u>Answer:</u> 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl
<u>Explanation:</u>
pH is defined as the negative logarithm of hydrogen ion concentration present in the solution
.....(1)
Given value of pH = 1.5
Putting values in equation 1:
![1.5=-\log[H^+]](https://tex.z-dn.net/?f=1.5%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{(-1.5)}=0.0316M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B%28-1.5%29%7D%3D0.0316M)
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
.....(2)
We are given:
Volume of solution = 15.0 mL
Molarity of HCl = 0.0316 M
Putting values in equation 2:

The chemical equation for the reaction of HCl and calcium carbonate follows:

By the stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of calcium carbonate
So,
of HCl will react with =
of calcium carbonate
The number of moles is defined as the ratio of the mass of a substance to its molar mass.

Moles of calcium carbonate = 
Molar mass of calcium carbonate = 100.01 g/mol
Putting values in the above equation:

Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl