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Answer: 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl
Explanation:
pH is defined as the negative logarithm of hydrogen ion concentration present in the solution
[tex]pH=-\log [H^+][/tex] .....(1)
Given value of pH = 1.5
Putting values in equation 1:
[tex]1.5=-\log[H^+][/tex]
[tex][H^+]=10^{(-1.5)}=0.0316M[/tex]
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:
[tex]\text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(2)
We are given:
Volume of solution = 15.0 mL
Molarity of HCl = 0.0316 M
Putting values in equation 2:
[tex]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[/tex]
The chemical equation for the reaction of HCl and calcium carbonate follows:
[tex]2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2[/tex]
By the stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of calcium carbonate
So, [tex]4.74\times 10^{-4}mol[/tex] of HCl will react with = [tex]\frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol[/tex] of calcium carbonate
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of calcium carbonate = [tex]2.37\times 10^{-4}mol[/tex]
Molar mass of calcium carbonate = 100.01 g/mol
Putting values in the above equation:
[tex]\text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g[/tex]
Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl
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