0.455 L (or 455 mL) of 1.60 x
M HCl is needed to neutralize 14.5 mL of a saturated
solution.
To determine the volume of 1.60 x
M HCl that is needed to neutralize 14.5 mL of a saturated
solution, we can first calculate the number of moles of
that are present in the solution.
We can do this by using the solubility of
in water (0.185 g/100.0 mL) and the volume of the solution (14.5 mL):
0.185 g
/ 100.0 mL * 14.5 mL = 0.026975 g
Next, we can use the molar mass of
(74.1 g/mol) to convert the mass of
to moles:
0.026975 g
/ 74.1 g/mol = 0.000364 mol
Since the reaction between HCl and
is a 2:1 mole ratio, we can then determine the number of moles of HCl that are needed to neutralize the
:
0.000364 mol Ca(OH)2 * 2 = 0.000728 mol HCl
Finally, we can use the concentration of the HCl solution (1.60 x
M) and the number of moles of HCl that are needed to determine the volume of HCl that is required:
0.000728 mol HCl / (1.60 x
M) = 0.455 L HCl
Therefore, 0.455 L (or 455 mL) of 1.60 x
M HCl is needed to neutralize 14.5 mL of a saturated
solution.