In an experiment, 20cm^3 of 0.2dm^3 NaOH reacted with 32.0cm^3 of H2SO4 solution. Calculate the concentration of the H2SO4 in a) mol dm^-3 and b) g dm^-3​

A. 0.0625 moldm¯³

B. 6.125 gdm¯³

Explanation:

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

2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O

From the balanced equation above,

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

The mole ratio of the base, NaOH (n₆) = 2

A. Determination of the concentration of the acid in moldm¯³.

Volume of base, NaOH (V₆) = 20 cm³

Concentration of base, NaOH (C₆) = 0.2 moldm¯³

Volume of acid, H₂SO₄ (Vₐ) = 32 cm³

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

The mole ratio of the base, NaOH (n₆) = 2

Concentration of acid, H₂SO₄ (Cₐ) =?

CₐVₐ / C₆V₆ = nₐ / n₆

Cₐ × 32 / 0.2 × 20 = 1 / 2

Cₐ × 32 / 4 = 1 / 2

Cross multiply

Cₐ × 32 × 2 = 4 × 1

Cₐ × 64 = 4

Divide both side by 64

Cₐ = 4 / 64

Cₐ = 0.0625 moldm¯³

Therefore, concentration of the acid in moldm¯³ is 0.0625 moldm¯³

B. Determination of the concentration of the acid in gdm¯³.

Concentration (moldm¯³) = 0.0625 moldm¯³

Concentration (gdm¯³) =?

Next, we shall determine the molar mass of the acid, H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (2×1) + 32 + (4×16)

= 2 + 32 + 98

= 98 g/mol

Finally, we shall determine the concentration of the acid in gdm¯³. This can be obtained as follow:

Concentration (moldm¯³) = 0.0625 moldm¯³

Molar mass of H₂SO₄ = 98 g/mol

Concentration (gdm¯³) =?

Conc. (moldm¯³) = conc. (gdm¯³) / Molar mass

0.0625 = Conc. (gdm¯³) / 98

Cross multiply

Conc. (gdm¯³) = 0.0625 × 98

Conc. (gdm¯³) = 6.125 gdm¯³

Thus, the concentration of the acid in gdm¯³ is 6.125 gdm¯³