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16 March, 13:07

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

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  1. 16 March, 13:50
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    Answer: 12.78ml

    Explanation:

    Given that:

    Volume of KOH Vb = ?

    Concentration of KOH Cb = 0.149 m

    Volume of HBr Va = 17.0 ml

    Concentration of HBr Ca = 0.112 m

    The equation is as follows

    HBr (aq) + KOH (aq) - -> KBr (aq) + H2O (l)

    and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

    Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va) / (Cb x Vb) = Na/Nb

    (0.112 x 17.0) / (0.149 x Vb) = 1/1

    (1.904) / (0.149Vb) = 1/1

    cross multiply

    1.904 x 1 = 0.149Vb x 1

    1.904 = 0.149Vb

    divide both sides by 0.149

    1.904/0.149 = 0.149Vb/0.149

    12.78ml = Vb

    Thus, 12.78 ml of potassium hydroxide solution is required.
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