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30 May, 10:34

At 100°C, the Kw of water is 5.6x10-13. What is the pOH of a solution that has a pH of 6.25 at 100°C? A) 6.00 B) 6.25 C) 7.00 D) 7.75 E) 8.00

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  1. 30 May, 12:19
    0
    The pOH of a solution that has a pH of 6.25 at 100°C is 7.75 (option D)

    Explanation:

    pOH + pH = 14

    pOH = 14 - 6.25

    pOH = 7.75

    Therefore, at 100°C, the Kw of water is 5.6x10⁻¹³ and the pOH of a solution that has a pH of 6.25 at 100°C is 7.75.

    Thus, option D, is the correct answer.
  2. 30 May, 13:48
    0
    D. 7.75.

    Explanation:

    Equation for the dissociation of water:

    H2O (aq) - -> H + (aq) + OH - (aq)

    [H+][OH-] = Kw

    The constant, Kw, is the ionic product of water. The product of concentrations of H + and OH - ions in water at a particular temperature is known as ionic product of water.

    Kw = 5.6x10-13

    Since,

    pH + pOH = 14

    pH = 6.25

    6.25 + pOH = 14

    Therefore,

    pOH = 14 - 6.25

    = 7.75.
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