Calculate the volume of 0.150 mol /L Ba(OH)2 necessary to neutralise 452 mL of 0.175 mol/L HCl.(First balance the neutralisation equation carefully.)

First, we need to balance the neutralization equation. The balanced equation for the reaction between barium hydroxide and hydrochloric acid is:

Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

From the balanced equation, we can see that one mole of Ba(OH)2 reacts with two moles of HCl.

Next, we calculate the number of moles of HCl. We can do this by multiplying the volume of the HCl solution (in liters) by its molarity.

Number of moles of HCl = volume (L) x molarity (mol/L) = 0.452 L x 0.175 mol/L = 0.0791 mol

Since one mole of Ba(OH)2 reacts with two moles of HCl, the number of moles of Ba(OH)2 needed is half the number of moles of HCl.

Number of moles of Ba(OH)2 = 0.0791 mol / 2 = 0.03955 mol

Finally, we calculate the volume of the Ba(OH)2 solution needed by dividing the number of moles of Ba(OH)2 by its molarity.

Volume of Ba(OH)2 = number of moles / molarity = 0.03955 mol / 0.150 mol/L = 0.264 L or 264 mL

So, 264 mL of 0.150 mol/L Ba(OH)2 is necessary to neutralize 452 mL of 0.175 mol/L HCl.