Question

How many grams of sulfuric acid are required to neutralize 125.0mL of 3.200M potassium hydroxide? What concentration of sulfuric acid would i need if i wanted to neutralize the base with only 50.00 mL.

Answer 1

The balanced equation is

2 KOH + H₂SO₄ --------> K₂SO₄ + 2 H₂O

Number of moles of KOH = molarity * volume of solution in L

Number of moles of KOH = 3.200 * 0.125 L = 0.400 mol

From the balanced equation we can say that

2 mole of KOH requires 1 mole of H2SO4 so

0.400 mole of KOH will require

= 0.400 mole of KOH *(1 mole of H2SO4 / 2 mole of KOH)

= 0.200 mole of H2SO4

number of moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4

0.200 mol = mass of H2SO4 / 98.079 g/mol

mass of H2SO4 = 19.6 g

Therefore, the mass of H2SO4 required would be 19.6 g

Coentration of H2SO4 = number of moles of H2SO4 / volume of solution in L

concentration of H2SO4 = 0.200 mol / 0.0500 L = 4.00 M

Therefore, the concentration of H2SO4 = 4.00 M