Question

What volume of 0.250 mol/L sulfuric acid, H2SO4(aq) is needed to react completely with 37.2 mL of 0.650 mol/L potassium hydroxide, KOH(aq)?

Answer 1

To find the volume of sulfuric acid solution needed, we can use stoichiometry and the balanced chemical equation for the reaction. By calculating the moles of KOH and then using the moles of H₂SO₄ needed to react completely with the given KOH solution, we can find the volume of the sulfuric acid solution. The volume of 0.250 M sulfuric acid solution needed is approximately 48.36 mL.

Step-by-step explanation:

To find the volume of 0.250 M sulfuric acid solution that will completely react with 37.2 mL of 0.650 M potassium hydroxide solution, we need to use stoichiometry and the balanced chemical equation for the reaction:

• H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

From the balanced equation, we can see that the mole ratio between H₂SO₄ and KOH is 1:2. Therefore, if we have 37.2 mL of 0.650 M KOH solution, we can calculate the moles of KOH:

• moles of KOH = volume (L) × concentration (M) = 0.0372 L × 0.650 mol/L = 0.02418 mol KOH

Using the stoichiometry of the reaction, we know that 1 mole of H₂SO₄ reacts with 2 moles of KOH. So, the moles of H₂SO₄ needed to react completely with the given KOH solution is:

• moles of H2SO4 = 0.02418 mol KOH × (1 mol H₂SO₄ / 2 mol KOH) = 0.01209 mol H₂SO₄

Finally, we can use the concentration of the sulfuric acid solution to calculate the volume:

• volume of H₂SO₄ = moles / concentration = 0.01209 mol / 0.250 mol/L = 0.04836 L = 48.36 mL

Answer 2

Answer: d. 48.4 ml

Step-by-step explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is
`H_2SO_4`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is KOH.

We are given:

`n_1=2\\M_1=0.250mol/L\\V_1=?mL\\n_2=1\\M_2=0.650mol/L\\V_2=37.2mL`

Putting values in above equation, we get:

`2* 0.250* V_1=1* 0.650* 37.2\\\\V_1=48.4ml`

Thus volume of 0.250 mol/L sulfuric acid needed is 48.4 ml