<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:
![0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol](https://tex.z-dn.net/?f=0.0316%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20HCl%7D%5Ctimes%201000%7D%7B15.0%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20HCl%7D%3D%5Cfrac%7B0.0316%5Ctimes%2015.0%7D%7B1000%7D%3D4.74%5Ctimes%2010%5E%7B-4%7Dmol)
The chemical equation for the reaction of HCl and calcium carbonate follows:
![2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2](https://tex.z-dn.net/?f=2HCl%2BCaCO_3%5Crightarrow%20H_2CO_3%2BCaCl_2)
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.
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Moles of calcium carbonate = ![2.37\times 10^{-4}mol](https://tex.z-dn.net/?f=2.37%5Ctimes%2010%5E%7B-4%7Dmol)
Molar mass of calcium carbonate = 100.01 g/mol
Putting values in the above equation:
![\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DCaCO_3%3D%282.37%5Ctimes%2010%5E%7B-4%7Dmol%29%5Ctimes%20100.01g%2Fmol%5C%5C%5C%5C%5Ctext%7BMass%20of%20%7DCaCO_3%3D0.0237g)
Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl