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10 September, 07:12

A 0.105 L sample of an unknown HNO 3 solution required 45.5 mL of 0.150 M Ba (OH) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

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  1. 10 September, 10:46
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    [HNO₃] = 0.0325 M

    Explanation:

    Formula for neutralization is:

    Normality of the acid. Acid volume = Normality of the base. Base volume

    We verify dissociations of base and acid

    HNO₃ → H⁺ + NO₃⁻ (Since we have 1 H⁺, N = M)

    Ba (OH) ₂ → Ba²⁺ + 2OH⁻ (We have 2 OH⁻, the N = M/2)

    [Ba (OH) ₂] = 0.150M / 2 = 0.075 N

    We replace dа ta:

    (First of all, we convert the volume of base from mL to L)

    45.5 mL. 1L / 1000mL = 0.0455 L

    0.105 L. Normality of the acid = 0.0455L. 0.075N

    Normality of the acid = (0.0455L. 0.075N) / 0.105L → 0.0325 N

    N = M → [HNO₃] = 0.0325 M
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