neutralizing balanced equations

Question

The balanced equation for the neutralization reaction of aqueous H2SO4H2SO4 with aqueous KOHKOH is shown.

H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)H2SO4(aq)+2KOH(aq)⟶2H2O(l)+K2SO4(aq)

What volume of 0.2500.250 M KOHKOH is needed to react completely with 10.610.6 mL of 0.2350.235 M H2SO4H2SO4?

Answer 1

Your question has many things duplicated. If the question should read: "what volume of 0.250 M KOH is needed to completely react with 10.6 mL of 0.235M H₂SO₄" then here is the answer.

You will need the balanced equation which gives that 2 moles of KOH reacts with 1 mol H₂SO₄. Given the volume and concentration (molarity of any solution you can calculate the moles of solute (reactant, in this case H₂SO₄. Using the mole ratio relationship from the balanced equation, you will be able to determine the number of moles of the 2nd reactant (KOH) which will be needed to react with it.

Remembering that when we are given the molarity of a solution it is telling the number of moles or the solute (KOH) present in 1 liter (1000 mL) of the solution, the volume of the solution can then be determined as follows:

(0.0106 L )(0.235 mol H₂SO₄ /L soln) ( 2 moles KOH/1 mol H₂SO₄) = 0.004982 mol KOH

Volume of KOH = 0.004982 mol KOH ) ( 1000 mL / 0.250 mol KOH ) = 19.9 mL

Explanation given above CAN be used to help solve ANY solution stoichiometry problem, not JUST this one. Hope this helps!