To determine the volume of 0.150 M HCl needed to neutralize 20.00 mL of 0.190 M Ba(OH)2, you can use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Ba(OH)2.
The balanced chemical equation is:
2 HCl + Ba(OH)2 -> BaCl2 + 2 H2O
From the equation, you can see that it takes two moles of HCl to react with one mole of Ba(OH)2.
First, calculate the number of moles of Ba(OH)2 in the given 20.00 mL of 0.190 M Ba(OH)2 solution:
Moles of Ba(OH)2 = concentration (M) x volume (L)
Moles of Ba(OH)2 = 0.190 M x 0.02000 L
Next, use the stoichiometry of the balanced equation to determine the moles of HCl required. Since the stoichiometric ratio is 2:1 for HCl to Ba(OH)2, the moles of HCl needed will be half of the moles of Ba(OH)2 calculated above.
Moles of HCl = (1/2) x Moles of Ba(OH)2
Finally, calculate the volume of 0.150 M HCl needed to have the calculated moles of HCl:
Volume of HCl (L) = Moles of HCl / concentration (M)
Volume of HCl = Moles of HCl / 0.150 M
Performing the calculations using the given values:
Moles of Ba(OH)2 = 0.190 M x 0.02000 L = 0.0038 moles
Moles of HCl = (1/2) x 0.0038 moles = 0.0019 moles
Volume of HCl = 0.0019 moles / 0.150 M = 0.0127 L or 12.7 mL
Therefore, approximately 12.7 mL of 0.150 M HCl is needed to neutralize 20.00 mL of 0.190 M Ba(OH)2.