Question

Sulfuric acid (H₂SO₄) reacts with potassium hydroxide (KOH) as follows:

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

Calculate the volume of 1.00M sulfuric acid required to neutralize 75mL of 0.100M KOH.

Answer 1

H₂SO₄(aq) + 2KOH(aq) ----------------------------> K₂SO₄(aq) + 2H₂O(l)

C1 = 1.00 M

V1 = ?

n1 = 1

C2 = 0.100 M

V2 = 75 mL

n2 =2

C1 V1 / n1 = C2 V2 / n2

1.00 x V1 / 1 = 0.100 x 75 / 2

V1 = 3.75 mL

volume of H2SO4 = 3.75 mL

Answer 2

H₂SO₄(aq) + 2KOH(aq) ----------------------------> K₂SO₄(aq) + 2H₂O(l)

C1 = 1.00 M

V1 = ?

n1 = 1

C2 = 0.100 M

V2 = 75 mL

n2 =2

C1 V1 / n1 = C2 V2 / n2

1.00 x V1 / 1 = 0.100 x 75 / 2

V1 = 3.75 mL

volume of H2SO4 = 3.75 mL