Question

The K_sp of Al(OH)₃ is 1.0 x 10^-33. What is the solubility of Al(OH)₃ in 0.000010 M NaOH? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).  

Answer 1

Following is the - complete Answer -&- Explanation, for the given Question, in.....typed format....

$\Rightarrow$Answer:

Molar solubility of Al(OH)3 , in 0.000010 M NaOH will be = 1.0 x 10 -18 M  ( mol/L )

$\Rightarrow$Explanation:

Following is the complete explanation, for the above Answer.....

• Given:

1. ​​​​​​​Value of solubility product constant:  Ksp = 1.0 x 10 -33 , of Al(OH)3 .
2. Molar concentration of sodium hydroxide: [NaOH] = 0.000010 M ( mol/L )

• ​​​​​​​Step - 1:

​​​​​​​We know, following is the balanced chemical equation, for ionic dissociation of Al(OH)3 , in its aqueous solution.

$\Rightarrow$balanced reaction: Al(OH)3 (aq)  $\rightleftharpoons$ Al3+ (aq) + 3 OH - (aq) ----------------------- Equation - (1)

$\Rightarrow$molar ratio:   Al(OH)3 : OH - = 1 : 3

$\Rightarrow$  molar ratio: Al(OH)3 : Al3+ = 1 : 1

• Step - 2:

​​​​​​​Therefore, following will be the expression of the solubility product constant ( Ksp ), of Aluminium hydroxide ( Al(OH)3 ):

$\Rightarrow$  Ksp =   [Al 3+] x [OH -] 3 = 1.0 x 10 -33    ------------------------------------------------- Equation - (2)

• Step - 3:

​​​​​​​We also know the following balanced chemical equation, of ionic dissociation of sodium hydroxide ( NaOH ), in its aqueous solution.

$\Rightarrow$   NaOH (aq)  $\rightleftharpoons$ Na+ (aq) + OH - (aq) ----------------------------------- Equation - ( 3 )

$\Rightarrow$ molar ratio: NaOH : OH - = 1 : 1

• Step - 4:

​​​​​​​Let's assume, at equilibrium, molar solubility of  Al(OH)3 ,is: [Al(OH)3] = S* ; i.e. [Al 3+]eq = S* and the molar solubility of OH - ion, arising from the dissociation of NaOH (aq), at equilibrium will be: [OH - ] = 0.000010 M ( mol/L)

$\Rightarrow$ Therefore, we can state that at equilibrium, following will be the molar solubility, of the Hydroxyl ion, ( OH - ):

$\Rightarrow$  [OH - ]eq = [  3.0 x ( S* ) +  0.000010 ]  M ( mol/L ) $\approx$ 0.000010 M  ( mol/L )

[ i.e. Considering, the value of molar solubiluity of Al(OH)3 , i.e. S* at equilibrium to be negligible, with respect to the molar concentration, of NaOH, at equilibrium. ]

• Step - 5:

​​​​​​​Therefore, plugging in values in Equation - ( 2 ), we will get the following:

$\Rightarrow$  Ksp =   [Al 3+]eq x [OH -]eq 3 = 1.0 x 10 -33

$\Rightarrow$ Ksp =   [ S* ] x [ 0.000010 ] 3 = 1.0 x 10 -33

$\Rightarrow$  S* =   ( 1.0 x 10 -33 ) / ( 0.000010 )3 =   1.0 x 10 -18  

• Step - 6 :

​​​​​​​Therefore, we have obtained the molar solubility, of Aluminium hydroxide, is the following:

$\Rightarrow$ Molar solubility of Al(OH)3 =  1.0 x 10 -18 M  ( mol/L )

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