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UNIT IV pH. I ONIZATION OF W ATER Pure distilled water still has a small conductivity. Why? 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq)

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Presentation on theme: "UNIT IV pH. I ONIZATION OF W ATER Pure distilled water still has a small conductivity. Why? 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq)"— Presentation transcript:

1 UNIT IV pH

2 I ONIZATION OF W ATER Pure distilled water still has a small conductivity. Why? 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq)

3 I ONIZATION OF W ATER 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq) In neutral water… [H 3 O + ] = [OH - ] In acidic solutions… [H 3 O + ] > [OH - ] In basic solutions… [OH - ] > [H 3 O + ] Keq = [H 3 O + ] [OH] or... Kw = [H 3 O + ] [OH - ]

4 I ONIZATION OF W ATER 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq) Since reaction is endothermic: At higher temperatures ______________are favoured and Kw is _____________er. At lower temperatures ______________ are favoured and Kw is _____________er.

5 I ONIZATION OF W ATER Always: [H 3 O + ] [OH - ] = Kw At 25 0 C only : [H 3 O + ] [OH - ] = 1.00 x

6 [H 3 O + ] & [OH - ] IN N EUTRAL W ATER At 25 o C: (NOTE: Assume T = 25 o C unless otherwise noted) [H 3 O + ] [OH - ] = 1.00 x and [H 3 O + ] = [OH - ] if water is neutral Substitute [H 3 O + ] for [OH - ]: [H 3 O + ] [H 3 O + ] = 1.00 x [H 3 O + ] 2 = 1.00 x [H 3 O + ] = 1.00 x = 1.00 x M Also [OH - ] = [H 3 O + ] = 1.00 x M

7 [H 3 O + ] & [OH - ] IN N EUTRAL W ATER Example Given: Kw at 60 0 C = 9.55 x Calculate [H 3 O + ] & [OH - ] at 60 0 C.

8 [H 3 O + ] & [OH - ] IN A CIDS AND B ASES 2H 2 O (l) H 3 O + (aq) + OH - (aq) Add acid, H 3 O + increases, so equilibrium shifts LEFT and [OH - ] decreases. Add base, [OH - ] increases, so the equilibrium shifts LEFT and [H 3 O + ] decreases.

9 [H 3 O + ] & [OH - ] IN A CIDS AND B ASES Ex. Find the [OH - ] in M HCl. Ex. Find [H 3 O + ] in M NaOH.

10 [H 3 O + ] & [OH - ] IN A CIDS AND B ASES Ex. Find [H 3 O + ] in M Ba(OH) 2. Ex. Calculate [OH - ] in M HNO 3 at 60 0 C. Kw at 60 0 C = 9.55 x Hebden Textbook Page 127 Questions #28-30

11 PHPH Shorthand method of showing acidity (or basicity/alkalinity) pH= -log[H 3 O + ] If [H 3 O + ] = 1.0 x pH = -log (1.0 x ) = 7

12 C ONVERTING [H 3 O + ] TO P H Ex. Find the pH of M HCl. **In pH or pOH notation, only the digits after the decimal are significant digits. The digits before the decimal come from the non-significant power of 10 in the original [H 3 O + ].

13 C ONVERTING [H 3 O + ] TO P H Ex. Find the pH of M NaOH at 25 0 C. Ex. Find the pH of neutral water at 25 0 C.

14 P H S CALE In neutral water pH = 7.0 In acid solution pH < 7.0 In basic solution pH > 7.0

15 C ONVERTING P H TO [H 3 O + ] [H 3 O + ] = antilog (-pH) Ex.) If pH = , find [H 3 O + ]. Ex.) If pH = calculate [H 3 O + ].

16 L OGARITHMIC N ATURE OF P H A change of 1 pH unit a factor of 10 in [H 3 O + ] (or acidity). How many times more acidic is pH 3 than pH 7? Natural rainwater pH ~ 6 Extremely acidic acid rain pH ~ 3 So, acid rain is 1000 times more acidic than natural rain water!

17 P OH pOH = -log [OH - ] [OH - ] = antilog (-pOH)

18 P OH Ex. Calculate the pOH of M KOH. Ex. Find the pH of the same solution. Notice: pH + pOH = 14.00

19 R ELATION OF P H TO P OH Since... [H 3 O + ] [OH - ] = Kw -log[H 3 O + ] + -log [OH - ] = -log Kw pH + pOH = pKw where pKw = -log Kw (definition of pKw) TRUE AT ALL TEMPERATURES!

20 R ELATION OF P H TO P OH Specifically at 25 0 C: Kw = 1.00 x pKw = -log (1.00 x ) pKw = pH + pOH = TRUE AT 25°C!

21 P H AND P OH Ex. Find the pH of 5.00 x M LiOH (25 0 C). Ex. Find the pOH of M HBr (25 0 C). See pOH scale & pH scale on page 140 of Hebden Textbook.

22 P H AND P OH When not at 25 0 C: Ex. At 60 0 C Kw = 9.55 x Find the pH of neutral water at 60 0 C.

23 P H AND P OH Is pH always 7.00 in neutral water?_________ At higher temperatures: 2H 2 O + heat H 3 O + + OH - [H 3 O + ] > 1.0 x so pH < 7 [OH - ] > 1.0 x so pOH < 7

24 S UMMARY OF P H AND P OH These are very important? Make sure you study these! In neutral water pH = pOH at any temperature pH & pOH = 7.00 at 25 0 C only At lower temps pH and pOH are > 7 At higher temps pH and pOH are < 7 At any temperature: pH + pOH = pKw At 25 0 C: pH + pOH = Hebden Textbook Pages Questions #49-53, 55-57

25 S UMMARY OF P H AND P OH Kw = [H 3 O + ][OH - ] [H 3 O + ] [OH - ] pH =-log[H 3 O + ] pOH = -log[OH - ] [H 3 O + ] = antilog(-pH) [OH - ] = antilog(-pH) pH pOH pH + pOH = 14


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