Solubility of Aluminum Hydroxide

Question

What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?

Answer 1

In order to determine the molar solubility of aluminum hydroxide, we must first determine the quantity of hydroxide in solution already. We were given a buffered solution of formic acid and sodium formate. So, we will have to proceed by an indirect method.

First, we will use the Henderson-Hasselbalch equation.

pH = pKa + log([HCOONa]/[HCOOH])

<=> 3.75 + log(0.1/01) <=> 3.75 + log(1) => pH = pKa = 3.75

Next, we want to know the [OH-]. So, we will use the pH we just calculated to figure it.

pH + pOH = 14 => pOH = 14 - pH

<=> 14 - 3.75

pOH = 10.25

pOH = -log[OH-]

10^-pH = [OH-]

<=> 10^^-10.25

[OH-] = 5.6x10^-11

Finally, we can set up an equilibrium equation.

Al(OH)_3(s) <----------> Al³⁺_(aq) + 3OH^-_(aq)

I ----- -------- 5.6x10^-11

Δ ----- +x +3x

Eq ----- x 5.6x10^-11 + 3x

K_sp = [Al³⁺][OH^-]³

1.9x10^-32 =x(5.6x10^-11 + 3x)³

=> 3x is negligible

1.9x10^-32 =x(5.6x10^-11)³

x = 1.9x10^-32/(5.6x10^-11)³

x = [Al³⁺] = 0.108M

So, the molar solubility of Al(OH)₃ in our formic acid/sodium formate buffer is 0.108M