The volume of 0.15 mol dm ^-3 sulphuric acid,H2SO4 required to completely neutralise 25.0cm^3 of potassium hydroxide,KOH solution.

To find the volume of sulfuric acid required to neutralize the potassium hydroxide solution, we can use the balanced equation:

H2SO4 + 2KOH -> K2SO4 + 2H2O

From the equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide to form 2 moles of water.

First, let's calculate the amount of potassium hydroxide in moles using the given volume and concentration:

Amount of KOH = Volume x Concentration
= 25.0 cm^3 x 0.15 mol/dm^3 (Remember to convert cm^3 to dm^3 by dividing by 1000)
= 0.025 dm^3 x 0.15 mol/dm^3
= 0.00375 mol

According to the stoichiometry of the balanced equation, 2 moles of KOH react with 1 mole of H2SO4. So, the amount of H2SO4 required will be half of the amount of KOH:

Amount of H2SO4 = 0.00375 mol / 2
= 0.001875 mol

Finally, we can calculate the volume of sulfuric acid needed using the volume and concentration:

Volume of H2SO4 = Amount / Concentration
= 0.001875 mol / 0.15 mol/dm^3
= 0.0125 dm^3 or 12.5 cm^3

Therefore, the volume of 0.15 mol/dm^3 sulfuric acid required to completely neutralize 25.0 cm^3 of potassium hydroxide solution is 12.5 cm^3.

To find the volume of the sulfuric acid required to completely neutralize the potassium hydroxide solution, we need to use the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH), which is as follows:

H2SO4 + 2 KOH → K2SO4 + 2 H2O

The equation shows that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide. However, the information given in the question is in terms of concentration and volume, so we need to convert the given information into moles.

First, let's calculate the number of moles of potassium hydroxide solution (KOH) using the given volume and concentration.

Number of moles of KOH = (Volume of KOH solution in dm^3) x (Concentration of KOH in mol/dm^3)
= (25.0 cm^3) x (1 dm^3 / 1000 cm^3) x (0.15 mol/dm^3)
= 0.00375 moles

From the balanced equation, we know that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide. Therefore, the number of moles of sulfuric acid required will be twice the number of moles of potassium hydroxide.

Number of moles of H2SO4 = 2 x Number of moles of KOH
= 2 x 0.00375 moles
= 0.0075 moles

Now, we need to find the volume of the sulfuric acid of 0.15 mol/dm^3 concentration required to contain 0.0075 moles. We can use the equation:

Volume of H2SO4 (in dm^3) = (Number of moles of H2SO4) / (Concentration of H2SO4 in mol/dm^3)
= 0.0075 moles / 0.15 mol/dm^3
= 0.05 dm^3

Therefore, the volume of 0.15 mol/dm^3 sulfuric acid required to completely neutralize 25.0 cm^3 of potassium hydroxide solution is 0.05 dm^3 (or 50.0 cm^3).