Calculate the pH of a saturated solution of iron(II) hydroxide. (See the appendi

Question

Calculate the pH of a saturated solution of iron(II) hydroxide. (See the appendix.)

Substance Ksp at 25 oC Aluminum Al(OH)3 1.9 x 10-33 Barium Ba(OH)2 5.0 x 10-3 BaCO3 8.1 x 10-9 BaSO4 1.1 x 10-10 Ba3(PO4)2 3.4 x 10-23 Cadmium Cd(OH)2 2.5 x 10-14 CdCO3 5.2 x 10-12 CdS 8.0 x 10-27 Calcium Ca(OH)2 5.5 x 10-6 CaCO3 4.8 x 10-9 Ca3(PO4)2 1.0 x 10-26 CaF2 3.9 x 10-11 Chromium Cr(OH)3 6.3 x 10-31 Cobalt Co(OH)2 1.6 x 10-15 CoS 4.0 x 10-21 Copper Cu(OH)2 2.2 x 10-20 Cu2S 1.6 x 10-48 CuCO3 1.4 x 10-10 CuS 6.3 x 10-36 Iron Fe(OH)2 8.0 x 10-16 FeS 6.3 x 10-18 Fe(OH)3 2.5 x 10-39 FePO4 9.9 x 10-29 Lead Pb(OH)2 2.8 x 10-16 PbF2 3.7 x 10-8 PbCl2 1.7 x 10-5 PbBr2 6.3 x 10-6 PbI2 6.5 x 10-9 PbCrO4 1.8 x 10-14 PbSO4 1.7 x 10-8 PbS 8.4 x 10-28 Magnesium Mg(OH)2 1.8 x 10-11 Manganese Mn(OH)2 1.9 x 10-13 MnCO3 8.8 x 10-11 MnS 5.6 x 10-16 Nickel Ni(OH)2 1.6 x 10-14 Silver AgCl 1.8 x 10-10 AgBr 5.0 x 10-13 AgI 8.3 x 10-17 AgCN 1.2 x 10-16 Ag2S 6.3 x 10-50 Ag2CrO4 1.1 x 10-12 Ag3PO4 2.6 x 10-18 Tin Sn(OH)2 1.4 x 10-28 SnS 1.3 x 10-23 Zinc ZnCO3 1.0 x 10-10 Zn(OH)2 4.5 x 10-17 ZnS 1.1 x 10-21

Explanation / Answer

given Fe(OH)2

Fe(OH)2 ---> Fe+2 + 2OH-

let the molar solubility of Fe(OH)2 be s

so

from the above equation we get

[Fe+2] = s

[OH-] = 2s

now

Ksp = [Fe+2] [OH-]^2

so

Ksp = [s] [2s]^2

Ksp = 4s3

from the given table

Ksp of Fe(OH)2 is 8 x 10-16

so

8 x 10-16 = 4s3

s = 5.848 x 10-6 M

so

[OH-] = 2s = 2 x 5.848 x 10-6

[OH-] = 1.1696 x 10-5 M

now

we know that

pOH = -log [OH-]

so

pOH = -log 1.1696 x 10-5

pOH = 4.93

now

pH = 14 - pOH

so

pH = 14 - 4.93

pH = 9.07

so

the pH of the given solution is 9.07

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