what volume in cm of 0.2 mil dm-3 hcl is required to neutralise 25cm3 of 0.2 mil dm-3 ba(oh)2

To solve this problem, we need to understand the concept of neutralization in chemistry. Neutralization is a reaction between an acid and a base which results in the formation of water (H2O) and a salt. The balanced chemical equation for the reaction between HCl (hydrochloric acid) and Ba(OH)2 (barium hydroxide) is:

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

From the balanced equation, we can see that 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2.

Given that the concentrations of both HCl and Ba(OH)2 are 0.2 mol/dm^3 and the volume of Ba(OH)2 is 25 cm^3, we can calculate the volume of HCl required.

First, convert the volume of Ba(OH)2 from cm^3 to dm^3:

25 cm^3 = 25/1000 dm^3 = 0.025 dm^3

Next, calculate the number of moles of Ba(OH)2 using the formula:

Number of moles = Concentration x Volume

Number of moles of Ba(OH)2 = 0.2 mol/dm^3 x 0.025 dm^3 = 0.005 mol

Since 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2, the number of moles of HCl required is:

Number of moles of HCl = 2 x Number of moles of Ba(OH)2 = 2 x 0.005 mol = 0.01 mol

Finally, calculate the volume of HCl required using the formula:

Volume = Number of moles / Concentration

Volume of HCl = 0.01 mol / 0.2 mol/dm^3 = 0.05 dm^3

Convert the volume from dm^3 to cm^3:

0.05 dm^3 = 0.05 x 1000 cm^3 = 50 cm^3

So, 50 cm^3 of 0.2 mol/dm^3 HCl is required to neutralize 25 cm^3 of 0.2 mol/dm^3 Ba(OH)2.