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3 February, 02:02

Which of the following aqueous solutions has the highest [OH - ]? A. A solution with a pH of 3.0 B. 1 * 10-4 M solution of HNO3 C. A solution with a pOH of 12.0 D. Pure water E. 1 * 10-3 M solution of NH4Cl

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  1. 3 February, 03:42
    0
    A solution with a pOH of 12.0

    Explanation:

    A. A solution with a pH of 3.0

    pH is 3 which is acidic or contain more of H+.

    B. 1 * 10-4 M solution of HNO3

    log of (1x10-4) = 4 still acidic (more of H+)

    C. A solution with a pOH of 12.0

    pOH = 12.0 has more of OH-

    D. Pure water

    has a balanced H + and OH-

    E. 1 * 10-3 M solution of NH4Cl

    log of (1x10-3) = 3 which is acidic, having more of OH-
  2. 3 February, 03:46
    0
    D. Pure water; [OH-] = 1e-7

    Explanation:

    The greater the concentration, the greater the basicity.

    A. A solution with a pH of 3.0

    pH + pOH = 14

    pOH = 14 - pH

    pOH = 14 - 3 = 11

    pOH = - log[OH-]

    [OH-] = Antilog (-pOH)

    [OH-] = Antilog (-11)

    [OH-] = 1e-11

    B. 1 * 10-4 M solution of HNO3

    pH = - log[H+]

    log10{0.0001} = log10 (10-4) = -4, by definition of the log function.

    Thus pH = 4,

    pH + pOH = 14

    pOH = 14 - pH

    pOH = 14 - 4 = 10

    pOH = - log[OH-]

    [OH-] = Antilog (-pOH)

    [OH-] = Antilog (-10)

    [OH-] = 1e-10

    C. A solution with a pOH of 12.0

    pOH = - log[OH-]

    [OH-] = Antilog (-pOH)

    [OH-] = Antilog (-12)

    [OH-] = 1e-12

    D. Pure water

    Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion concentration.

    [OH-] = 1e-7

    E. 1 * 10-3 M solution of NH4Cl

    NH4Cl dissolves in solution to form ammonium ions NH+4 which act as a weak acid by protonating water to form ammonia, NH3 (aq) and hydronium ions H3O + (aq):

    NH+4 (aq) + H2O (l) → NH3 (aq) + H3O + (aq)

    Ka * Kb = 1.0*10-14 assuming standard conditions.

    So, Ka (NH+4) = 1.0*10-14 / 1.8*10-5 = 5.56*10-10

    Plug in the concentration and the Ka value into the expression:

    Ka = [H3O+] * [NH3] / [NH+4]

    5.56*10-10 ≈ [H3O+] * [NH3] / [0.001]

    5.56*10-13 = [H3O+]^2

    (as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, Ka is small, so x≪0.1.)

    [H3O+] = 7.46*10-7

    pH=-log[H3O+]

    pH = - log (7.45*10-6)

    pH ≈ 6.13

    pH + pOH = 14

    pOH = 14 - pH

    pOH = 14 - 6.13 = 7.87

    pOH = - log[OH-]

    [OH-] = Antilog (-pOH)

    [OH-] = Antilog (-7.87)

    [OH-] = 1.345e-08
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