Based on the titration results, what is the concentration of the sulfuric acid?

Volume of 𝐾𝑂𝐻 36 mL
Concentration of 𝐾𝑂𝐻 0.16 M
Volume of 𝐻2𝑆𝑂4 12 mL

H2SO4 + 2KOH ==> K2SO4 + 2H2O

mols KOH = M x L = ?
Look at the equation to see that mols H2SO4 = 1/2 mols KOH
Then M H2SO4 = mols/L
Post your work if you get stuck.

Hsb

To find the concentration of sulfuric acid (H2SO4) based on the titration results, you can use the principle of stoichiometry. In a neutralization reaction between potassium hydroxide (KOH) and sulfuric acid, the mole ratio is 1:2. This means that for every mole of KOH, two moles of H2SO4 react.

Let's start by calculating the number of moles of KOH:

Moles of KOH = Volume of KOH (in liters) Γ— Concentration of KOH (in mol/L)

Given that the volume of KOH is 36 mL and the concentration is 0.16 M, we need to convert the volume to liters:

Volume of KOH = 36 mL Γ— (1 L/1000 mL) = 0.036 L

Now we can calculate the moles of KOH:

Moles of KOH = 0.036 L Γ— 0.16 mol/L = 0.00576 mol

Since the mole ratio between KOH and H2SO4 is 1:2, the moles of H2SO4 will be twice the moles of KOH:

Moles of H2SO4 = 2 Γ— Moles of KOH

Moles of H2SO4 = 2 Γ— 0.00576 mol = 0.01152 mol

To find the concentration of the sulfuric acid, divide the moles of H2SO4 by the volume of H2SO4 in liters:

Concentration of H2SO4 = Moles of H2SO4 / Volume of H2SO4 (in liters)

Given that the volume of H2SO4 is 12 mL, which is equivalent to 0.012 L:

Concentration of H2SO4 = 0.01152 mol / 0.012 L = 0.96 M

Therefore, the concentration of the sulfuric acid is 0.96 M based on the titration results.