Determine the volume of 0.250 M KOH solution required to neutralize each of the following samples of sulfuric acid. The neutralization reaction is:
H2SO4(aq) + 2KOH(aq) = K2SO49aq) + 2H2O(l)
Find: 15 mL of 0.250 M H2SO4
To determine the volume of 0.250 M KOH solution required to neutralize 15 mL of 0.250 M H2SO4, we can use the stoichiometry of the neutralization reaction.
1. Start by writing down the balanced equation for the neutralization reaction:
H2SO4(aq) + 2KOH(aq) = K2SO4(aq) + 2H2O(l)
2. Apply the stoichiometry of the balanced equation to determine the ratio of reactants involved. From the balanced equation, we see that 1 mole of H2SO4 reacts with 2 moles of KOH.
3. Calculate the number of moles of H2SO4 in the 15 mL of 0.250 M H2SO4:
moles H2SO4 = volume (in L) × concentration (in mol/L)
moles H2SO4 = (15 mL ÷ 1000 mL/L) × 0.250 mol/L
4. Use the mole ratio from the balanced equation to determine the number of moles of KOH required for neutralization. According to the balanced equation, it takes 2 moles of KOH to neutralize 1 mole of H2SO4.
moles KOH = moles H2SO4 × (2 moles KOH / 1 mole H2SO4)
5. Finally, calculate the volume of 0.250 M KOH solution required to neutralize the given amount of H2SO4. Use the definition of concentration:
volume of KOH solution (in L) = moles KOH / concentration (in mol/L)
Plug in the values and calculate:
volume of KOH solution = moles KOH / 0.250 mol/L
Make sure to convert the volume from liters to milliliters if needed.
By following these steps, you should be able to calculate the volume of 0.250 M KOH solution required to neutralize 15 mL of 0.250 M H2SO4.