Test#1 Problem#5

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CosetteBackus_4F
Posts: 23
Joined: Wed Feb 20, 2019 12:18 am

Test#1 Problem#5

Postby CosetteBackus_4F » Tue Feb 11, 2020 1:30 pm

Can someone explain how to get the pOH of the solution after finding the Ka?

The pKb of fluoride is 10.8. A 0.12M solution of hydrogen fluoride was made for silicon wafer processing. What is the pOH of this solution at equilibrium?
HF(aq) + H2O(l) -> F-(aq) + H3O+(aq)

Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Re: Test#1 Problem#5

Postby Jessica Booth 2F » Tue Feb 11, 2020 2:02 pm

After finding the Ka set up your ice table. You should get that the equilibrium conc. are [HClO2]= .15-x, [ClO2-]=x and [H3O+]=x. Then set up your equilibrium equation with the appropriate values. The Kc should = x^2/(.15-x) which also equals the Ka that is .012. Then solve the quadratic and you get that the [H3O+]=.036849 M. From here you can either find the pH and then subtract the pH from 14 to find the pOH or use the fact that [H3O+][OH-]= kW=10^-14 and then take the -log of [OH-]. The final pOH should be 12.57.


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