1 / 12

pOH

pOH. What is the [H + ]?. If the [H + ] = 1 x 10 -7 M, the pH = 7. That works well with acids, but what about bases? Consider sodium hydroxide: NaOH(aq)  Na + + OH - What is the [H + ]?. We need a new formula!!. Since bases release hyroxides, OH - , we will calculate the pOH.

genica
Download Presentation

pOH

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. pOH

  2. 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+]?

  3. 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.

  4. 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

  5. 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!

  6. 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

  7. 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.

  8. 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 + 1.7 = 14 pH = 12.3

  9. 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.

  10. 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[0.00045] = 3.35 • pH + 3.35 = 14 • pH = 10.65

  11. 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 + 1.14 = 14 • pH = 12.86

  12. 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.

More Related