Given that the enthalpy of neutralization for the reaction of HCl (a strong acid) and NaOH (a strong base) is always -55.90 kJ per mole of H2O formed, what is the concentration of a 75 mL sample of HCl if the enthalpy of neutralization for the reaction was found to be -6.21 kJ, assuming no heat is lost to the calorimeter?
To find the concentration of the HCl solution, we need to use the given information about the enthalpy of neutralization and the volume of the solution.
First, let's determine the number of moles of water (H2O) formed in the reaction using the enthalpy of neutralization.
Enthalpy of neutralization = -55.90 kJ/mol of H2O
Enthalpy of neutralization for the given reaction = -6.21 kJ
We can set up a proportion to find the moles of water formed:
-55.90 kJ / mol H2O = -6.21 kJ / x
Cross-multiplying gives:
-55.90 kJ * x = -6.21 kJ * mol H2O
x = (-6.21 kJ * mol H2O) / (-55.90 kJ)
x ≈ 0.111 mol H2O
Since the reaction between HCl and NaOH has a 1:1 stoichiometry, the number of moles of water formed is equal to the number of moles of HCl.
Next, we need to calculate the molarity (concentration) of the HCl solution.
Molarity (M) = moles of solute / volume of solution (in liters)
Volume of solution = 75 mL = 0.075 L
Molarity (M) = 0.111 mol HCl / 0.075 L
M ≈ 1.48 M
Therefore, the concentration of the 75 mL sample of HCl is approximately 1.48 M.
To determine the concentration of HCl in the 75 mL sample, we need to use the enthalpy of neutralization and the given data.
The enthalpy change for the reaction of HCl and NaOH is always -55.90 kJ per mole of water formed. We can use this information to figure out the number of moles of water produced in the reaction.
Given that the enthalpy change for the reaction in question is -6.21 kJ and assuming no heat loss to the calorimeter, we can set up the following equation:
-6.21 kJ = n * (-55.90 kJ/mol)
where n is the number of moles of water formed.
Rearranging the equation to solve for n, we have:
n = -6.21 kJ / (-55.90 kJ/mol)
n ≈ 0.111 mol
Since 1 mole of HCl reacts with 1 mole of water, the number of moles of HCl present in the 75 mL sample is also approximately 0.111 mol.
To find the concentration, we can use the formula:
concentration = moles / volume
In this case, the volume is given as 75 mL, which can be converted to liters by dividing by 1000:
volume = 75 mL / 1000 = 0.075 L
Now we can calculate the concentration:
concentration = 0.111 mol / 0.075 L
concentration ≈ 1.48 mol/L
Therefore, the concentration of the 75 mL sample of HCl is approximately 1.48 mol/L.
55.90 kJ/mol x # mol = 6.21 kJ. Solve for # mols.
Then M = #mols/L. You know # mols and L (0.075 L). Calculate M.