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    PH, pOH, PKw

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    Calculating pH, POH, PKa, PKb.

    37) PH 0.600 M solution of Na2S
    Sodium sulfide is a salt of a weak acid (hydrogen sulfide) and a strong alkali (sodium hydroxide). It therefore undergoes hydrolysis in water.
    Na2S = 2Na+ + S2-
    S2- + 2H2O = H2S + 2OH-
    Kh = [H2S][OH-]
    [S2-]
    [OH-] = Kh[S2-]
    [H2S]
    POH = -logKh + log[H2S] - log[S2-]
    PKw = PH + POH
    POH = 14 - PH
    14 - PH = PKh + log[H2S] - log[S2-]
    PH = 14 - PKh + log [S2-] - log[H2S]
    S2- + 2H2O = H2S + 2OH-
    T=0 1 mol 0 0
    T=t (1-x) mol x mol 2x mol
    Conc. c(1-x) cx 2cx

    Kh = [H][OH-]
    [CN-]
    = [HCN](Kw/[H+])
    [CN-]
    = [HCN]Kw
    [H+][CN-]
    [H+][CN-] = Kw/Kh
    [HCN]
    [H+][CN-] = Ka
    [HCN]
    Ka = Kw
    Kh
    Kh = Kw
    Ka
    Kh = cx.2cx
    c(1-x)
    1-x is negligible, therefore,
    Kh = 2cx2
    x = (Kh/2c)1/2
    = (Kw/Ka.c)1/2
    = (10-14/Ka x 2x6x10-1)1/2
    =(10-13/12Ka)1/2
    cx = [OH-]
    = 6 x10-1 x (10-13/12Ka)1/2
    POH= -log [6 x10-1 x (10-13/12Ka)1/2]
    = 1 - log6 + ½(13 + log 12 + log Ka)
    = 1 - 0.7782 + 6.5 + 1.0792 + log Ka
    = 7.801 + log Ka
    PH = pKw - pOH
    = (14 - 7.801) - log Ka
    = PKa + 6.199
    Substitute pKa for H2S.

    © BrainMass Inc. brainmass.com October 9, 2019, 6:55 pm ad1c9bdddf
    https://brainmass.com/chemistry/physical-chemistry/ph-poh-pkw-102472

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    37) PH 0.600 M solution of Na2S
    Sodium sulfide is a salt of a weak acid (hydrogen sulfide) and a strong alkali (sodium hydroxide). It therefore undergoes hydrolysis in water.
    Na2S = 2Na+ + S2-
    S2- + 2H2O = H2S + 2OH-
    Kh = [H2S][OH-]
    [S2-]
    [OH-] = ...

    Solution Summary

    PH, pOH, PKw calculations are provided. Sodium sulfide as a weak acid is determined.

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