Question

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Answer 1

The solubility product constant (K_sp) of Al(OH)₃ can be equated as follows.

Al(OH)₃(s) <----> Al³⁺(aq) + 3OH^-(aq); K_sp = 3*10^-34

The formation constant (K_f) of Al(OH)₄^- can be equated as follows.

Al³⁺(aq) + 4OH^-(aq) <----> Al(OH)₄^-(aq); K_f = 2*10³³

If you add the above two equations, the corresponding equilibrium constants can be written as the product, as shown below.

Al(OH)₃(s) + OH^-(aq) <----> Al(OH)₄^-(aq); K = K_sp * K_f = (3*10^-34) * (2*10³³) = 0.6

The relation between pH and pOH can be written as follows.

pH + pOH = 14

i.e. pOH = 14 - pH

Now, pOH can be written as follows.

pOH = -Log[OH-]

i.e. [OH-] = 10^-pOH