Question

​given that ksp=1.6x10^-14 for iron hydroxide fe(oh) calculate the molar solubility of this compound in a buffer solution at 9.0 ph

Answer 1

Ksp is given for Fe(OH) but it should be Fe(OH)₂. I have solved for ferrous hydroxide.

Answer 2

To calculate the molar solubility of iron hydroxide (Fe(OH)2) in a buffer solution at pH 9.0, we need to consider the hydrolysis reaction of Fe(OH)2 in water.

The hydrolysis reaction of Fe(OH)2 can be represented as follows: Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH-(aq)

In a buffer solution at pH 9.0, the concentration of hydroxide ions (OH-) can be calculated using the equation: [OH-] = 10^(-pOH)

Given that pH = 9.0, we can calculate pOH as: pOH = 14 - pH = 14 - 9 = 5.0

Substituting the value of pOH into the equation, we find: [OH-] = 10^(-5.0) = 1 x 10^(-5) M

Since the hydrolysis reaction of Fe(OH)2 involves the formation of 2 OH- ions for every Fe2+ ion, the molar solubility of Fe(OH)2 can be determined by dividing the concentration of OH- ions by 2: Molar solubility of Fe(OH)2 = [OH-] / 2 = (1 x 10^(-5)) / 2 = 5 x 10^(-6) M

Therefore, the molar solubility of iron hydroxide (Fe(OH)2) in a buffer solution at pH 9.0 is 5 x 10^(-6) M.