pH, pOH, PKw

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 + Ѕ(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.