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15 August, 23:27

An aqueous solution of hydroiodic acid is standardized by titration with a 0.168 M solution of sodium hydroxide. If 18.7 mL of base are required to neutralize 29.6 mL of the acid, what is the molarity of the hydroiodic acid solution

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  1. 16 August, 00:07
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    0.106M

    Explanation:

    Equation of reaction

    NaOH + HI = ==" NaI + H₂O

    If 0.168 moles of NaOH = 1000cm³

    X moles = 18.7 cm³

    X = (18.7 * 0.168) / 1000 = 0.00314 moles

    From equation of reaction,

    1 moles of NaOH = 1 mole of HI

    0.00314 moles of NaOH = 0.00314 moles of HI

    Therefore,

    0.00314 moles of HI = 29.6cm³

    X moles = 1000cm³

    X = (0.00314 * 1000) / 29.6 = 0.106M
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