Question

Solid potassium hydroxide is slowly added to 175 mL of a 0.0487 M aluminum acetate solution. The concentration of hydroxide ion required to just initiate precipitation is M.

Answer 1

Solution

The aluminium acetate reacts with potassium hydroxide to form aluminium hydroxide and is precipitated out.

Now, the dissociation of Aluminium hydroxide is as

Al(OH)₃ ⇌ Al³⁺ + 3OH¯

K_sp expression is as

K_sp = [Al³⁺] [OH¯]³

K_sp of Al(OH)₃ is 1.90 x 10^-33

[Al³⁺] = 0.0487 M

[OH¯] = ?

Now, plug into the K_sp expression

1.90 x 10^-33 = (0.0487)(s)³

s = [(1.90 x 10^-33)/ (0.0487)]^1/3 = 3.39 x 10^-11 M

Concentration of hydroxide ion, [OH¯], required to just initiate precipitation = 3.39 x 10^-11 M