Question

A solution of sulfuric acid (H2SO4, 25.00 mL) was titrated to completion with 34.55 mL of 0.1020 M sodium hydroxide. What was the concentration of the sulfuric acid

Answer 1

Answer:

C=0.140964M

Explanation:

M1V1=M2V2

DATA

M1=?

M2=0.1020M

V1=25.00ML

V2=34.55ML

M*25.00ML=(0.1020M)*(34.55ML)

M*25.00ML=3.5241M*ML

M=3.5241M*ML/25.00ML

M=0.140964M

Therefore, concentration of sulfuric acif is 0.140964M

Answer 2

Answer:

0.0705 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂SO₄ (nA) = 1

The mole ratio of the base, NaOH (nB) = 2

Next, data obtained from the question. This include the following:

Volume of acid, H₂SO₄ (Va) = 25 mL

Volume of base, NaOH (Vb) = 34.55 mL

Molarity of base, NaOH (Mb) = 0.1020 M

Molarity of acid, H₂SO₄ (Ma) =?

The molarity of the acid, H₂SO₄ can be obtained as follow:

MaVa / MbVb = nA/nB

Ma × 25 / 0.1020 × 34.55 = 1/2

Ma × 25 / 3.5241 = 1/2

Cross multiply

Ma × 25 × 2 = 3.5241

Ma × 50 = 3.5241

Divide both side by 50

Ma = 3.5241 / 50

Ma = 0.0705 M

Therefore, the concentration of the acid, H₂SO₄ is 0.0705 M