You are given 5.00 ml of an h2so4 solution of unknown concentration. you divide the 5.00-ml sample into five 1.00-ml samples and titrate each separately with 0.1000 m naoh. in each titration, the h2so4 is completely neutralized. the average volume of naoh solution used to reach the endpoint is 15.3 ml. what was the concentration of h2so4 in the 5.00-ml sample

The concentration of H₂SO₄ = 0.765 M

Explanation:

Given: Concentration of NaOH: M₁ = 0.10 M, Volume of NaOH: V₁ = 15.3 mL, Concentration of H₂SO₄: M₂ = ?, Volume of H₂SO₄: V₂ = 1 mL

Reaction involved: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

In this titration reaction, one mole H₂SO₄ neutralizes two moles NaOH.

To find out the concentration of H₂SO₄, we use the equation: N₁ ·V₁ = N₂ ·V₂

Here, N₁, N₂ are the normality of NaOH and H₂SO₄, respectively.

For NaOH: normality: N₁ = M₁

For a diprotic acid like H₂SO₄: normality: N₂ = 2 × M₂

∴ N₁ ·V₁ = N₂ ·V₂

⇒ 0.10 M × 15.3 mL = [2 × M₂] × 1 mL

⇒ M₂ = (0.10 M × 15.3 mL) ÷ (2 × 1 mL) = 0.765 M

Therefore, the concentration of H₂SO₄ = 0.765 M

answer

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