# Problem: Calculate the pH of a 0.050-M Al(NO 3)3 solution. The Ka value for Al(H2O)63+ is 1.4 X 10-5.

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###### FREE Expert Solution

$\overline{){\mathbf{pH}}{\mathbf{=}}{\mathbf{-}}{\mathbf{log}}{\mathbf{\left[}}{{\mathbf{H}}}_{{\mathbf{3}}}{{\mathbf{O}}}^{{\mathbf{+}}}{\mathbf{\right]}}}$

Species in solution: Al(H2O)63+, NO3-, H2O

• Al(H2O)63+Stronger acid than H2O

Dominant equilibriumAl(H2O)63+(aq) + H2O(l)  H3O+(aq) +Al(H2O)52+(aq)

ICE table: Ka expression:

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}\mathbf{Al}{{\mathbf{\left(}\mathbf{H}}_{}}^{}}{}}$2O)52+][Al(H2O)63+]

Solving for x = [H3O+] ###### Problem Details

Calculate the pH of a 0.050-M Al(NO 3)3 solution. The Ka value for Al(H2O)63+ is 1.4 X 10-5.