What volume of 0.1244M hydrochloric acid is required to neutralize 24.81 ml of 0.1026M barium hydroxide?
from the mole to mole ratio and the given information I came up with .00509 mols of HCl. would I multiply this by 0.1244 or divide by .1244. I'm not sure which would be correct. I know one way mols will cancel out and leave me with liters, but not sure which one
I don't get the part where you got the .00509 mol HCl.
This is how I solved it.
Ba(OH)2 + 2HCl ===> BaCl2 + 2H2O
Note:
M (molarity) = moles of solute / liters of solution
m (molality) = moles of solute / kg of solvent
24.81 ml --> 0.02481 L
Tip: to make ur life easier, set the conversions in a table.
Volume of HCl:
(1L HCL / 0.1244 mol HCl)*(2 mol HCl / 1 mol Ba(OH)2)*(0.1026 mol Ba(OH)2 / 1L Ba(OH)2)*(0.02481L Ba(OH)2) = 0.0409L HCl