25cm^3 H2SO4 is neutralized by 40 cm^3 of 0.03 mol/dm^3 KOH. What is the molar concentration of H2SO4?
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols KOH = M x L = ?
Using the coefficients in the balanced equation, convert mols KOH to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4.
40cm^3 * (1dm/10cm)^3 * .03mol/dm^3 = 0.0012 mol KOH
Each mole of H2SO4 needs 2 moles of KOH to be neutralized.
That means you had .0006 mol H2SO4
.0006mol/25cm^3 * 1000cm^3/L = 0.24mol/L = 0.24M H2SO4
To find the molar concentration of H2SO4, we need to use the volume and concentration of KOH, as well as the reaction stoichiometry between H2SO4 and KOH.
The balanced equation for the neutralization reaction between H2SO4 and KOH is as follows:
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the balanced 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 react completely.
Given that the volume of KOH used is 40 cm^3, we can convert it to dm^3 by dividing by 1000:
40 cm^3 = 40/1000 dm^3 = 0.04 dm^3
The concentration of KOH is given as 0.03 mol/dm^3.
Now, let's calculate the number of moles of KOH used in the neutralization reaction:
moles of KOH = concentration × volume
moles of KOH = 0.03 mol/dm^3 × 0.04 dm^3
moles of KOH = 0.0012 mol
Since the molar ratio between H2SO4 and KOH is 1:2, the number of moles of H2SO4 used in the neutralization reaction is half the number of moles of KOH used:
moles of H2SO4 = 0.0012 mol / 2
moles of H2SO4 = 0.0006 mol
The volume of H2SO4 used is given as 25 cm^3. Again, we can convert it to dm^3:
25 cm^3 = 25/1000 dm^3 = 0.025 dm^3
Finally, we can calculate the molar concentration of H2SO4:
molar concentration of H2SO4 = moles of H2SO4 / volume of H2SO4
molar concentration of H2SO4 = 0.0006 mol / 0.025 dm^3
molar concentration of H2SO4 = 0.024 mol/dm^3
Therefore, the molar concentration of H2SO4 is 0.024 mol/dm^3.