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6 March, 06:55

What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?

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  1. 6 March, 10:28
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    Answer is: volume of H₂SO₄ is 42.1 mL.

    Chemical reaction: H ₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

    c (H ₂SO₄) = 0,4567 M = 0,4567 mol/L.

    V (NaOH) = 30 mL : 1000 mL/L = 0,03 L.

    c (NaOH) = 0,321 M = 0,321 mol/L.

    n (NaOH) = c (NaOH) · V (NaOH).

    n (NaOH) = 0,321 mol/L · 0,030 L.

    n (NaOH) = 0,00963 mol.

    From chemical reaction: n (H ₂SO₄) : n (NaOH) = 1 : 2.

    n (H ₂SO₄) = 0,01926 mol.

    V (H ₂SO₄) = n (H₂SO₄) : c (H₂SO₄).

    V (H ₂SO₄) = 0,01926 mol : 0,4567 mol/L.

    V (H ₂SO₄) = 0,0421 L = 42,1 mL.
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