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.

🤓 Based on our data, we think this question is relevant for Professor Cole & Geng & Lovander's class at IOWA.

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pH=-log[H3O+]



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:


Ka=productsreactants=[H3O+][Al(H2O)52+][Al(H2O)63+]


Solving for x = [H3O+]



Ka=1.4×10-5=[H3O+][Al(H2O)52+][Al(H2O)63+]1.4×10-5=[x][x][0.050 - x], ignore -x:[Al(H2O)63+]iKa>5001.4×10-5(0.050)=x2[0.050](0.050)7.0×10-7=x2


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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.

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Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.

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Based on our data, we think this problem is relevant for Professor Cole & Geng & Lovander's class at IOWA.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.