Assuming that 0.5 grams of a 1.0 gram sample is CaCO3, what should the concentration of HCl(M) be if you want to use 25.0 mL for the titration?
What indicator are you using for the titration. Phenolphthalein titrates it half way; methyl orange titrates all of it.
we're using methyl orange
To find the concentration of HCl (in moles per liter, or M) required for the titration, we need to use stoichiometry and the concept of balanced chemical equations.
First, let's write the balanced chemical equation for the reaction between CaCO3 (calcium carbonate) and HCl (hydrochloric acid):
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
From the balanced equation, we can see that one mole of CaCO3 reacts with two moles of HCl. Thus, we can use the stoichiometry to establish the relationship between the number of moles of CaCO3 and the number of moles of HCl.
To start, we need to calculate the number of moles of CaCO3 in the given sample. We are told that 0.5 grams of the 1.0-gram sample is CaCO3. The molar mass of CaCO3 is approximately 100.09 g/mol.
Moles of CaCO3 = (Mass of CaCO3 in grams) / (Molar mass of CaCO3)
= 0.5 g / 100.09 g/mol
= 0.004997 mol (rounded to five decimal places)
Now, we know that the number of moles of CaCO3 is equal to the number of moles of HCl since they react in a 1:2 ratio (1 mole CaCO3: 2 moles HCl).
Moles of HCl = 0.004997 mol (CaCO3) * (2 mol HCl) / (1 mol CaCO3)
= 0.009994 mol (rounded to five decimal places)
To find the concentration of HCl, we can use the following equation:
Molarity (M) = (Number of moles of solute) / (Volume of solution in liters)
We are given that the volume used for titration is 25.0 mL, which is equivalent to 0.0250 L.
Concentration of HCl (M) = (0.009994 mol) / (0.0250 L)
= 0.39976 M (rounded to five decimal places)
Therefore, the concentration of HCl should be approximately 0.400 M if you want to use 25.0 mL for the titration.