6B.5 (c) asks to calculate the pH and pOH of this strong base: 0.0092 M Ba(OH)2
When I find the negative log of 0.0092 to find the pOH, I get 2.036, but the answer key says 1.74.
What am I missing?
6B.5
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Megan Wang 3D
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- Joined: Mon Jan 09, 2023 9:55 am
Re: 6B.5
For Ba(OH)2, there's two OH- ions for every Ba(OH)2 that dissociates. So you would have the multiply the concentration of Ba(OH)2 by 2 to get the OH- concentration before calculating the pOH and pH.
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