6D.15

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Rita Chen 1B
Posts: 112
Joined: Sat Jul 20, 2019 12:15 am

6D.15

Postby Rita Chen 1B » Mon Jan 20, 2020 11:28 pm

How do we calculate the pH of 0.055 M AlCl3 (aq)?

Benjamin Feng 1B
Posts: 102
Joined: Sat Sep 07, 2019 12:19 am

Re: 6D.15

Postby Benjamin Feng 1B » Mon Jan 20, 2020 11:38 pm

You have to look at the ions created: Al3+ and Cl-. Cl- is the conjugate for a strong acid so it does not contribute to the pH. But, Al3- can form Al(OH)3, and some will form. You have to write the Kb equation to see how the concentration of OH- changes to find the final pOH and therefore the pH as well. Kb = [OH-]^3[Al3+]/[Al(OH)3]

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: 6D.15

Postby nicolely2F » Mon Jan 20, 2020 11:42 pm

Use the ICE table. The initial concentration is just 0.055M of AlCl3, the rest is 0. Then, there will be a change of -x in [AlCl3] and +x in each of the products. The equilibrium concentration will be 0.055-x, x, x. Plug these expressions into the Ka expression and solve the resulting expression to discover x (x being [H3O+]). Then calculate the pH using pH = -log[H3O+]


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