What volume of 2.5*10^-2 M H2SO4 is required to neutralize 125mL if 0.03 M KOH
changing concentrations to Normality: 2.5e-2 M= 5e-2N on sulfuric
.03M KoH=.03N KOH
Na*Va=Nb*Vb
Va=(.03/5e-2)125ml=75ml
To find the volume of 2.5 * 10^-2 M H2SO4 required to neutralize 125 mL of 0.03 M KOH, we need to use the concept of stoichiometry. This involves using the balanced chemical equation between H2SO4 and KOH to determine the mole ratios.
The balanced equation for the neutralization reaction between H2SO4 and KOH is:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
From the equation, we can see that the mole ratio between H2SO4 and KOH is 1:2. This means that for every 1 mole of H2SO4, we need 2 moles of KOH to neutralize.
First, let's calculate the number of moles of KOH in the given volume of 125 mL:
moles of KOH = volume (L) * concentration (M)
moles of KOH = 125 mL * 0.03 M
moles of KOH = 3.75 mmol (since 1 mL = 1 mmol when the concentration is given in M)
Next, using the mole ratio, we know that we need 2 moles of KOH to react with 1 mole of H2SO4. Therefore, we need half the number of moles of KOH to react with the H2SO4.
moles of H2SO4 = 0.5 * moles of KOH
moles of H2SO4 = 0.5 * 3.75 mmol
moles of H2SO4 = 1.875 mmol
Now, let's calculate the volume of H2SO4 required using the given concentration of 2.5 * 10^-2 M:
volume of H2SO4 (L) = moles / concentration
volume of H2SO4 (L) = 1.875 mmol / 2.5 * 10^-2 M
volume of H2SO4 (L) = 0.075 L (since 1 mmol = 1 mL when the concentration is given in M)
volume of H2SO4 (L) = 75 mL
Therefore, 75 mL of the 2.5 * 10^-2 M H2SO4 is required to neutralize 125 mL of 0.03 M KOH.