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3 March, 22:41

A 0.1873 g sample of a pure, solid acid, h2x was dissolved in water and titrated with 0.1052 m naoh solution. the balanced equation for the neutralization reaction occurring is h2x (aq) + 2naoh (aq)  na2x (aq) + 2h2o (l) if the molar mass of h2x is 85.00 g/mol, calculate the volume of naoh solution needed in the titration.

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  1. 4 March, 01:18
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    According to the balanced equation of the reaction:

    H2X (aq) + 2 NaOH (aq) → Na2X (aq) + 2H2O (l)

    First, we have to get the no. of moles of H2X:

    no. of moles of H2X = weight / molar mass

    when we have the H2X weight = 0.1873 g & the molar mass H2X = 85 g/mol

    So by substitution:

    ∴ no. of moles of H2X = 0.1873 / 85

    = 0.0022 mol

    -then, we need to get no. of moles of NaOH:

    from the balanced equation, we can see that 1 mol H2X = 2 mol NaOH

    ∴ no. of moles of NaOH = no. of moles of H2X * 2

    = 0.0022 * 2 = 0.0044 mol

    So we can get the volume per litre from this formula:

    M (NaOH) = no. of moles NaOH / Volume L

    So by substitution:

    0.1052 = 0.0044 / Volume L

    ∴Volume = 0.042 L * 1000 = 42 mL
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