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pOH
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What is the [H+]? If the [H+] = 1 x 10-7M, the pH = 7.
That works well with acids, but what about bases? Consider sodium hydroxide: NaOH(aq) Na+ + OH- What is the [H+]?
![What is the [H+] If the [H+] = 1 x 10-7M, the pH = 7.](http://slideplayer.com/slide/14726836/90/images/2/What+is+the+%5BH%2B%5D+If+the+%5BH%2B%5D+%3D+1+x+10-7M%2C+the+pH+%3D+7..jpg "That works well with acids, but what about bases Consider sodium hydroxide: NaOH(aq) Na+ + OH- What is the [H+]")
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Since bases release hyroxides, OH-, we will calculate the pOH.
We need a new formula!! Since bases release hyroxides, OH-, we will calculate the pOH. pOH = -log[OH-] The formula is very similar to the pH formula, in fact you use the calulator the same way.
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Let’s consider water Water is H2O, but let’s write it HOH.
Some water molecules naturally dissociate: HOH H+ + OH- In fact, 1 x 10-7M of the water molecules dissociate When this happens, you end up with: [H+] = 1 x 10-7M and the [OH-] = 1 x 10-7M
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What is the pH? Therefore we can calculate that water has a pH of 7.
Using the new formula pOH = -log[OH-]; we can calculate that water has a pOH of 7. Therefore, pH + pOH = 14. This works for all solutions, not just water!
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Who’s on First? Let’s consider our first base, NaOH. What is the pOH of a 0.02M NaOH(aq) solution? NaOH(aq) Na+ + OH- All of the NaOH will break apart (dissociate), therefore the [Na+] = [OH-] = 0.02M
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pOH = -log[OH-] Now use the formula: pOH = -log[0.02] pOH = 1.7
If you are having trouble using your calculator, ask your teacher for help.
![pOH = -log[OH-] Now use the formula: pOH = -log[0.02] pOH = 1.7](http://slideplayer.com/slide/14726836/90/images/7/pOH+%3D+-log%5BOH-%5D+Now+use+the+formula%3A+pOH+%3D+-log%5B0.02%5D+pOH+%3D+1.7.jpg "If you are having trouble using your calculator, ask your teacher for help.")
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But what is pOH? Outside of chemistry class, pOH is not used very much, but pH is used often. So, what is the pH of 0.02M NaOH(aq)? pH + pOH = 14 pH = 14 pH = 12.3
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Does that make sense? Should a 0.02M NaOH(aq) solution have a pH of 12.3? Sure!! Remember that solutions with a pH greater than 7 are basic and NaOH(aq) is a basic solution.
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Practice What is the pH of a 0.00045M KOH(aq) solution?
Work out the answer using your calculator and then click to see the answer. pOH = -log[ ] = 3.35 pH = 14 pH = 10.65
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Practice Again What is the pH of a 0.073M LiOH(aq) solution?
Work out the answer using your calculator and then click to see the answer. pOH = -log[0.073] = 1.14 pH = 14 pH = 12.86
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One more practice problem
If a CuOH(aq) solution has a pH of 9.5, what is the concentration (molarity) of the CuOH(aq)? Work out the answer using your calculator and then click to see the answer. 9.5 + pOH = 14, therefore the pOH = 4.5 4.5 = -log[OH-], therefore [OH-] = 3.16 x 10-5 Since one CuOH gives off one OH-, the [CuOH] = 3.16 x 10-5M.
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