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Yesterday’s Homework Page 611 # 19 Page 612 # 20
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [H + ] = 1.0 x 10 -2 log [H + ] = -2 pH = 2
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [H + ] = 3.0 x 10 -6 log [H + ] = 0.477 – 6 = -5.523 pH = 5.523
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [OH - ] = 8.2 x 10 -6 log [OH - ] = 0.913 – 6 = -5.087 pOH = 5.087
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Calculate the pH of solutions having the following ion concentrations at 298 o K. pOH = 5.087 pH + pOH = 14 pH = 14 – pOH = 14 – 5.087 = 8.913
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [OH - ] = 1.0 x 10 -6 log [OH - ] = -6 pOH = 6 pH = 14 – 6 = 8
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [OH - ] = 6.5 x 10 -4 log [OH - ] = 0.812 - 4 pOH = 3.188 pH = 14 – 3.188 = 10.812
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Calculate the pH of solutions having the following ion concentrations at 298 o K. [H + ] = 3.6 x 10 -9 log [H + ] = 0.556 - 9 pH = 8.444 pOH = 14 – 8.444 = 5.556
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[H + ] =.025 Calculate the pH of solutions having the following ion concentrations at 298 o K.
![[H + ] =.025 Calculate the pH of solutions having the following ion concentrations at 298 o K.](http://images.slideplayer.com./30/9545519/slides/slide_9.jpg "[H + ] =.025 Calculate the pH of solutions having the following ion concentrations at 298 o K.")
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[H + ] = 2.5 x 10 -2 log [H + ] = 0.397 – 2 pH = 1.603 pOH = 14 – 1.603 = 12.397 Calculate the pH of solutions having the following ion concentrations at 298 o K.
![[H + ] = 2.5 x log [H + ] = – 2 pH = pOH = 14 – = Calculate the pH of solutions having the following ion concentrations at 298 o K.](http://images.slideplayer.com./30/9545519/slides/slide_10.jpg "[H + ] = 2.5 x log [H + ] = – 2 pH = pOH = 14 – = Calculate the pH of solutions having the following ion concentrations at 298 o K.")
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Now let’s try it the other way around…
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Given the pH, calculate the [H + ] and [OH - ] for the following solution. pH = 7.40 -log [H + ] = 7.40 log [H + ] = -7.40 =.60 – 8 [H + ] = 4.0 x 10 -8
![Given the pH, calculate the [H + ] and [OH - ] for the following solution.](http://images.slideplayer.com./30/9545519/slides/slide_12.jpg "pH = log [H + ] = 7.40 log [H + ] = =.60 – 8 [H + ] = 4.0 x")
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Given the pH, calculate the [H + ] and [OH - ] for the following solution. pOH = 14 – 7.40 = 6.60 log [OH - ] = -6.60 =.40 – 7 [OH - ] = 2.5 x 10 -7
![Given the pH, calculate the [H + ] and [OH - ] for the following solution.](http://images.slideplayer.com./30/9545519/slides/slide_13.jpg "pOH = 14 – 7.40 = 6.60 log [OH - ] = =.40 – 7 [OH - ] = 2.5 x")
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One last thing… If you have a solution of 1.0 M HCl, what is the [H + ]?
![One last thing… If you have a solution of 1.0 M HCl, what is the [H + ]](http://images.slideplayer.com./30/9545519/slides/slide_14.jpg "One last thing… If you have a solution of 1.0 M HCl, what is the [H + ]")
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One last thing… HCl is a strong acid. That means it dissociates completely. If [HCl] = 1.0, then [H+] = 1.0
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One last thing… Same with strong bases. If [NaOH] = 1.0, then [OH - ] = 1.0
![One last thing… Same with strong bases. If [NaOH] = 1.0, then [OH - ] = 1.0](http://images.slideplayer.com./30/9545519/slides/slide_16.jpg "One last thing… Same with strong bases. If [NaOH] = 1.0, then [OH - ] = 1.0")
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Homework Page 614 Problems 21 & 22
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