What volume of 0.15 mom hclis required to neutralize 25ml of 0.12molar ba(oh)2 solution?
(25/1000) .12 (2) = 0.006 Mols of OH-
so need .006 mols of H+
.006 mols/ (.15 mols/Liter)= .04 Liters
= 40 mL
To determine the volume of 0.15 M HCl required to neutralize 25 mL of 0.12 M Ba(OH)2 solution, we can use the concept of neutralization and stoichiometry.
First, let's write down the balanced chemical equation for the reaction between HCl and Ba(OH)2:
2HCl + Ba(OH)2 --> 2H2O + BaCl2
From the equation, we can see that it takes two moles of HCl to react with one mole of Ba(OH)2. This means that the stoichiometric ratio between HCl and Ba(OH)2 is 2:1.
Now, we can calculate the number of moles of Ba(OH)2 present in 25 mL of 0.12 M solution:
moles of Ba(OH)2 = concentration (M) x volume (L)
= 0.12 mol/L x 0.025 L
= 0.003 mol
Since the stoichiometric ratio is 2:1, we would need twice the number of moles of HCl to react:
moles of HCl needed = 2 x 0.003 mol
= 0.006 mol
Finally, we can calculate the volume of 0.15 M HCl required to provide 0.006 moles:
volume of HCl = moles of HCl / concentration (M)
= 0.006 mol / 0.15 mol/L
= 0.04 L or 40 mL
Therefore, 40 mL of 0.15 M HCl is required to neutralize 25 mL of 0.12 M Ba(OH)2 solution.