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22 April, 19:12

A) Calculate the pH of a 1.8 * 10 - 4 M HCl solution

b) Calculate the [H + ] and pOH of a 0.05 M KOH solution

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  1. 22 April, 19:53
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    Answer:-

    A) 3.745

    B) 2 x 10^-13 M [H+]

    pOH = 1.3

    Explanation:-

    A) From the question we see

    The strength of HCl solution = 1.8 x 10^-4 M.

    The M stands for molarity which is moles / Litre.

    HCl is a monobasic acid. So the number of moles of Hydrogen ion H + HCl can give is the same as the value of it's strength in moles per litre.

    The hydrogen ion concentration [H+] is therefore 1.8 x 10^-4.

    The formula for pH is

    pH = - log [ H + ]

    = - log [1.8 x 10^-4]

    = 3.745

    B) From the question we see

    The strength of KOH = 0.05 M

    The M stands for molarity which is moles / Litre.

    KOH is a monoacidic base. So the number of moles of Hydroxyl ion OH - KOH can give is the same as the value of it's strength in moles per litre.

    The Hydroxyl ion concentration [OH-] is therefore 0.05

    The relation between [OH-] and [H+] is

    [H+] x [ OH-] = 10^-14

    [H+] = 10^-14 / [OH-]

    = 10^-14 / 0.05

    = 2 x 10^-13 M

    Using the relation between pH and [OH-]

    pOH = - log [OH-]

    = - log [0.05]

    = 1.3
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