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: 205 mL of 0.130 M H2SO4
Bleh!
To determine the volume of the KOH solution required to neutralize the given sample of H2SO4, we need to use the balanced chemical equation provided.
The balanced equation is:
H2SO4(aq) + 2KOH(aq) = K2SO4(aq) + 2H2O(l)
By examining the equation, we can see that the molar ratio between H2SO4 and KOH is 1:2. This means that for every 1 mole of H2SO4, we need 2 moles of KOH to achieve a complete neutralization.
To find the number of moles of H2SO4 in the given sample, we'll use the formula:
moles = molarity × volume (in liters)
Given:
Molarity (H2SO4) = 0.130 M
Volume (H2SO4) = 205 mL = 0.205 L
Using the formula, moles of H2SO4 = 0.130 M × 0.205 L = 0.02665 moles
Since the molar ratio between H2SO4 and KOH is 1:2, we need twice the moles of KOH to neutralize the acid.
moles KOH = 2 × moles H2SO4 = 2 × 0.02665 moles = 0.0533 moles
Now, let's determine the volume of the KOH solution required. We'll use the same formula as before:
moles = molarity × volume (in liters)
Given:
Molarity (KOH) = 0.250 M
Volume (KOH) = ?
Rearranging the formula, we get:
Volume (KOH) = moles / molarity
Plugging in the values,
Volume (KOH) = 0.0533 moles / 0.250 M = 0.213 L
To convert the volume from liters to milliliters:
Volume (KOH) = 0.213 L × 1000 mL/L = 213 mL
Therefore, 213 mL of 0.250 M KOH solution is required to neutralize 205 mL of 0.130 M H2SO4.