To find the volume of 1.5 M HCl required to neutralize 18 mL of 2.0 M KOH, we can use the equation:
M1V1 = M2V2
where M1 is the molarity of the first solution (HCl), V1 is the volume of the first solution we need to find, M2 is the molarity of the second solution (KOH), and V2 is the volume of the second solution we have (18 mL).
First, let's find the number of moles of KOH in 18 mL of 2.0 M KOH:
moles KOH = M2 x V2
moles KOH = 2.0 mol/L x 0.018 L
moles KOH = 0.036 mol
Since HCl and KOH react in a 1:1 ratio, we need 0.036 moles of HCl to completely neutralize the KOH.
Now we can use the equation M1V1 = M2V2 to solve for V1:
M1V1 = M2V2
(1.5 mol/L) V1 = (2.0 mol/L) (0.018 L)
V1 = (2.0 mol/L) (0.018 L) / (1.5 mol/L)
V1 = 0.024 L or 24 mL
Therefore, 24 mL of 1.5 M HCl is required to completely neutralize 18 mL of 2.0 M KOH.