Question

In the neutralization reaction

H₂SO₄ + 2 KOH ⇌⇌ 2 H₂O + K₂SO₄

If there is 846 mL of 0.153 M H₂SO₄ reacting with 541 mL of a KOH solution with unknown concentration, what is the concentration of KOH?

Answer 1

Chemical equation must be balanced. Equation given in question is already balanced.

From balanced equation we understand that 2 moles of KOH needed to 1 mole of H₂SO₄ for neutralization.

Given - Conc. of H₂SO₄ = 0.153 M

vol. of H₂SO₄ = 846 ml= 0.846 L

vol. of KOH = 541 ml = 0.541 L

To find - = conc. of KOH (say X)

Formula - [ Molarity (M) = No. of moles of solute / liter of solution. ]

Solution - Using this formula we can find moles of H₂SO₄ and KOH.

Therefore,

NO. of moles of H₂SO₄ = 0.153 x 0.846

= 0.129 moles.

moles of KOH must be twice of moles of H₂SO₄ i.e 0.259 moles.

NO. of moles of KOH = X x 0.541

0.259 moles = X x 0.541

X   = 0.259/ 0.541

= 0.478

Therefore, 541 ml of 0.478 M KOH  needed to neutralize 846 ml 0f 0.531 M H₂SO₄.