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Compute and compare the molar solublities for aluminum hydroxide, Al(OH)3, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.

Sagot :

b)

Concentartion of hydroxide ion:

pOH = 14 - 4.74

pOH = 9.26

[OH-] = 10^-9.26 = 5.5x10^-10

Therefore: The solubility of Al(OH)3 in this buffer is then calculated from its solubility product expressions:

[tex]K_{sp}=\lbrack Al^{3+}^{}\rbrack\lbrack OH^-\rbrack^3[/tex]

The molar solubility in buffer = [Al^3+] = Ksp/[OH-] = 2x10^-32

2x10^-32 = Al (5.5x10^-10)^3

Al = 1.2x10^-4 M

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