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27 October, 19:09

What quantity of 18.68 M KOH solution is needed to neutralize 0.2841 grams of KHP? What is the concentration of the KOH solution?

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  1. 27 October, 19:31
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    KHP stands for potassium hydrogen phthalate which is the monopotassium salt of phthalic acid.

    KHP dissociates completely in water to form K (+) + HP (-); and HP (-).

    HP (-) dissolves as per HP (-) + H2O ⇄ P (2-) + H3O (+).

    That means that KHP the neutralization ratio of KOH and KPH is 1:1.

    So, 1 mole of KOH neutralizes 1 mol of KPH.

    Calculate the moles of KPH in 0.2841 grams.

    You need the molar mass of KPH. Its molecular formula is C8H5KO4 and its molar mass is 204.22 g/mol

    = n = 0.2841 g / 204.22 g/mol = 0.00139 moles of KPH.

    Now, calculate the volumen of 18.68 M KOH solution that contains that number of moles:

    M = n / V = > V = n / M = 0.00139 moles / 18.68 M = 0.0000744 l = 0.0744 ml.

    The concentration of the KOH solution is given in the statement: 18.68 M.

    That is a weird value for the concentration so may be your question has some mistake.

    Answer: 0.0744 ml
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