To neutralize 8.34 ml of HCl 36.5% by mass and density =
1, 2 g/cm3, the required volume of 5 mol/L NaOH will be?
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ?
Then mols HCl = M x L = ?
mols NaOH = mols HCl
Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.
Awesome ! thanks!@ Drbob
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ?
That line should read
1.2 g/cc x 1000 cc x 0.365 x (1/36.5) = ?
It's the .65 that is not right. It should be 0.365 and that comes from the 36.5%
To find the required volume of 5 mol/L NaOH needed to neutralize 8.34 ml of HCl 36.5% by mass, we first need to determine the number of moles of HCl.
1. Calculate the mass of HCl:
Mass = Volume * Density
Mass = 8.34 ml * 1.2 g/cm^3 (since density = 1.2 g/cm^3)
Mass = 10.008 g
2. Calculate the number of moles of HCl:
Number of moles = Mass / Molar mass
The molar mass of HCl is 36.5 g/mol.
Number of moles = 10.008 g / 36.5 g/mol
Number of moles ≈ 0.2744 moles
Now, to neutralize 1 mole of HCl, we require 1 mole of NaOH. Since the concentration of NaOH is given as 5 mol/L, we can calculate the volume of NaOH needed.
3. Volume of NaOH = Number of moles of HCl * Volume of NaOH per mole of HCl
Volume of NaOH = 0.2744 moles * (1 L / 5 mol)
Volume of NaOH = 0.05488 L (or 54.88 ml)
Therefore, the required volume of 5 mol/L NaOH to neutralize 8.34 ml of HCl 36.5% by mass and density 1.2 g/cm^3 is approximately 54.88 ml.