Question

Find the pH of a 0.008 M solution of an aluminum compound which dissolves completely to give the hydrated aluminum ion Al(H2O)3+6.

Al(H2O)3+6(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)5(OH)2+(aq),    Ka=1.4×10−5

• Report your answer with three significant figures.

  Find the pH of a 0.008 M solution of an aluminum compound which dissolves completely to give the hydrated aluminum ion Al(H2O)3+6.

  Al(H2O)3+6(aq)+H2O(l)⇌H3O+(aq)+Al(H2O)5(OH)2+(aq),    Ka=1.4×10−5

• Report your answer with three significant figures.

Answer 1

Given:-

molar concentration of [Al(H₂O)₆]³⁺ = 0.008 M

equilibrium constant of [Al(H₂O)₆]³⁺ = 1.4 $\times$ 10^-5

As we know that 0.008 M solution of aluminum compound dissolves completely to give the hydrated aluminum ion i.e [Al(H₂O)₆]³⁺ then molar concentration of [Al(H₂O)₆]³⁺ would be 0.008 M and then equilibrium established between [Al(H₂O)₆]³⁺ and solution of aluminum compound i.e[Al(H₂O)₅ (OH)]²⁺ which is given as follows:-

[Al(H₂O)₆]³⁺ + H₂O_(l)$\rightleftharpoons$ [Al(H₂O)₅ (OH)]²⁺_(aq) + H₃O⁺_(aq)

0.008 M 0 0

(0.008 - X) X X

As we know that

equilibrium constant of [Al(H₂O)₆]³⁺ (Ka) = [ [Al(H₂O)₅ (OH)]²⁺ ][H₃O⁺] / [ [Al(H₂O)₆]³⁺ ][H₂O]

equilibrium constant of [Al(H₂O)₆]³⁺ (Ka) =X $\times$X / (0.008 - X) $\times$ 1 ( since [H₂O] = 1)

neglecting the X in denominator in above expression then we get

equilibrium constant of [Al(H₂O)₆]³⁺ (Ka) =  X² / 0.008

1.4 $\times$ 10^-5 =  X² / 0.008

X² = 0.008 $\times$  1.4 $\times$ 10^-5

X² = 0.0112$\times$ 10^-5

X² = 11.2$\times$ 10^-8

X = $\sqrt{}$11.2$\times$ 10^-8

X   = 3.3466 $\times$ 10^-4

therefore

molar concentration of [H₃O⁺] = 3.3466 $\times$ 10^-4

molar concentration of [ [Al(H₂O)₅ (OH)]²⁺ ]= 3.3466 $\times$ 10^-4

As we know that

pH = - log[H₃O⁺]

pH = - log[3.3466 $\times$ 10^-4 ]

pH = - [log(3.3466) + ( -4log(10) ]

pH = - [(0.5246) - ( 4) ]

pH = 4.0000 - 0.5246

pH = 3.4754 (i.e the answer)