Tutorial 3: Ionization of Water

Transcription

1 Tutorial 3: Ionization of Water DEMONSTRATION OF CONDUCTIVITY OF TAP WATER AND DISTILLED WATER - Pure distilled water still has a small conductivity. Why? - There are a few ions present. - Almost all the pure water is H 2 O molecules. - But every once in a while, this happens: H + H - O + H H O + O H H O H H A proton is transferred Hydronium Hydroxide From one water molecule to another. An equilibrium is established Equation: Ionization of Water H 2 O + H 2 O H 3 O + + OH - Or 2H 2 O (l) H 3 O + (aq) + OH - (aq) Process is Endothermic 2H 2 O (l) + 59KJ H 3 O + (aq) + OH - (aq) All water or aqueous solutions contain these. In neutral water [H 3 O + ] = [OH - ] In acidic solutions [H 3 O + ] > [OH - ] Know these!! In basic solutions [OH - ] > [H 3 O + ] Write the Keq Expression for this equilibrium: Keq = 59KJ + 2H 2 O (l) H 3 O + (aq) + OH - (aq) Keq = [H 3 O + ] [OH] (liquid water left out) Given a special name for ionization of water called Kw So Kw = [H 3 O + ] [OH - ] Ionization constant for water Always true at any temp! Since reaction is endothermic: 59KJ + 2H 2 O (l) H 3 O + (aq) + OH - (aq)

2 At higher temps are favoured and Kw is er. At lower temps are favoured and Kw is er. At 25 o C (only) Kw = 1.00 x Know this!! For comparison At 10 0 C Kw = x (smaller) At 60 0 C Kw = 9.55 x (larger) So Always: [H 3 O + ] [OH - ] = Kw At 25 0 C only: [H 3 O + ] [OH - ] = 1.00 x [H 3 O + ] & [OH - ] in Neutral Water At 25 o C (NOTE: Assume Temp = 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 - ]) (If water is mentioned in a problem, it can be assumed to be NEUTRAL unless otherwise stated!) [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 10-7 M Also [OH - ] = [H 3 O + ] = 1.00 x 10-7 M At Higher Temp Given: Kw at 60 0 C = 9.55 x Calculate [H 3 O + ] & [OH - ] at 60 0 C [H 3 O + ] & [OH - ] in Acids and Bases 2H 2 O (l) H 3 O + (aq) + OH - (aq) Add acid, H 3 O + increases, so equilibrium shifts LEFT and [OH - ] decreases All acids produce H 3 O + in water Add base, [OH - ] increases, so the equilibrium shifts LEFT and [H 3 O + ] decreases. Finding [H 3 O + ] and [OH - ] in Acids and Bases

3 At 25 0 C Eg.) Find the [OH - ] in M HCl In a Strong Acid ([H 3 O + ] = [acid]) [H 3 O + ] = M [H 3 O + ][OH - ] = 1.00 x [OH - ] = 1.00 x = 1.00 x = 1.00 x M [H 3 O + ] 1.00 x 10-2 Find [H 3 O + ] in M NaOH. [OH - ] is less in an acid than in neutral water. [H 3 O + ][OH - ] = 1.00 x [H 3 O + ] = 1.00 x = 1.00 x = 3.33 x [OH - ] [H 3 O + ] is very low in a strong fairly concentrated base Find [H 3 O + ] in M Ba(OH) 2 In a STRONG BASE, [OH - ] = [base] x # of OH s [OH - ] =? ( ) M [H 3 O + ] = 1.00 x = M ( ) At Other Temps - you d be given Kw eg.) Kw at 60 0 C = 9.55 x Calculate [OH - ] in M HNO 3 at 60 0 C. [H 3 O + ][OH - ] = Kw SA [H 3 O + ][OH - ] = 9.55 x [OH - ] = 9.55 x = 1.59 x M ph -Shorthand method of showing acidity (or basicity, alkalinity) If [H 3 O + ] = 0.10 M (1.0 x 10-1 M) ph = 1.00

4 [H 3 O + ] = M (1.0 x10-4 M) ph = 4.00 Definition of ph ph= -log 10 [H 3 O + ] (assume log = log 10 ) If [H 3 O + ] = 1.0 x 10-7 ph = -log (1.0 x 10-7 ) Find the ph of M HCl 2 SD s SA. So [H 3 O + ] = [acid] [H 3 O + ] = M ph= -log (0.030) = How to round off?? Sig. Digits in ph start at decimal point!!! so ph = 1.52 Start counting here. Find the ph of neutral water at 25 0 C [H 3 O + ] = 1.00 x 10-7 ph = Find the ph of M NaOH at 25 0 C [H 3 O + ] = 1.00 x = 1.00 x M No units for ph Basic solution [OH - ] At 25 o C so ph = sig. digits In neutral water ph = 7.0 In acid solution ph < 7.0 In basic solution ph > 7.0

5 ph Scale 25 o C) More Acidic Neutral More Basic Calculate ph of 12.0 M HCl Very concentrated Acid! Answer Another example: Calculate the ph of 15.0 M NaOH: [H 3 O + ] = 1.00 x = 6.67 x M 15.0 Very concentrated Base ph = -log (6.67 x ) ph = Converting ph to [H 3 O + ] ph = - log [H 3 O + ] -ph = log [H 3 O + ] antilog (-ph) = [H 3 O + ] or [H 3 O + ] = antilog (-ph) 3 SD s eg.) If ph = , find [H 3 O + ] Put in (-) before you take the antilog [H 3 O + ] = antilog ( ) For regular Scientific Calculator: Enter: /- 2 nd log The calculator answer should be x The original ph had 3 SD s, so the answer must also have 3 SD s (Remember the answer is NOT a ph, so digits to the LEFT of the decimal point are also significant!). Remember that concentration also has a unit!. So the answer would be reported as: [H 3 O + ] = 2.44 x M 3 SD s unit For a DAL or TI 83 enter: 2 nd log (-) (or +/-) ENTER (or =) If ph = calculate [H 3 O + ]

6 Logarithmic Nature of ph A change of 1 ph unit a factor of 10 in [H 3 O + ] or (acidity) eg.) ph = 3.0 [H 3 O + ] = 1 x 10-3 M ph goes 10 times more acidic down 1 unit ph = 2.0 [H 3 O + ] = 1 x 10-2 M How many times more acidic is ph 3 than ph 7? ph 7 [H 3 O + ] = 1 x 10-7 x 10 4 = 10,000x ph 3 [H 3 O + ] = 1 x 10-3 or taking antilog of difference in ph 7 3 = 4 antilog 4 = 10 4 = 10,000 times (remember lower ph more acidic) Natural rainwater ph ~ 6 Extremely acidic acid rain ph ~ 3 diff = 3 & antilog (3) = 10 3 (1,000) So, the acid rain is 1000 times more acidic than natural rain water! poh = -log [OH - ] poh And [OH - ] = antilog (-poh) Calculate the poh of M KOH [OH - ] = 2.0 x 10-3 M poh = -log (2.0 x 10-3 ) = 2.70 Find the ph of the same solution: [OH - ] = 2.0 x 10-3 M [H 3 O + ] = 1.00 x = 5.0 x x 10-3 ph = Notice: ph + poh = From Math: If a x b = c Then: log a + log b = log c Eg.) 10 x 100 = 1000 Log(10) + log(100) = log(1000) = 3

7 So since [H 3 O + ] [OH - ] = Kw log[h 3 O + ] + log[oh - ] = log (Kw) or make everything negative -log[h 3 O + ] + -log [OH - ] = -log Kw ph + poh = pkw (relation) True at all temperatures where pkw = -log Kw (definition of pkw) Specifically at 25 0 C Kw = 1.00 x pkw = -log (1.00 x ) pkw = Only at 25 0 C so at 25 0 C ph + poh = At 25 0 C If ph = 4.00 poh = Or: If ph = poh= eg.) Find the ph of 5.00 x 10-4 M LiOH (25 0 C) plan: [OH - ] poh ph [OH - ] = 5.00 x 10-4 poh = ph = = eg.) Find the poh of M HBr (25 0 C) [H 3 O + ] = M (HBr is a strong acid) ph = poh = poh = When not at 25 0 C Eg.) At 60 0 C Kw = 9.55 x Find the ph of neutral water at 60 0 C. One way: Calculate pkw pkw = -log Kw = -log (9.55 x ) At 60 0 C pkw =

8 For neutral water ph = poh ([H 3 O + ] = [OH - ]) ph + poh = pkw (substitute ph for poh) ph + ph = pH = so ph = = Is ph always 7.00 in neutral water? At higher temp: 2H 2 O + heat H 3 O + + OH - Summary: [H 3 O + ] > 1.0 x 10-7 so ph < 7 [OH - ] > 1.0 x 10-7 so poh < 7 In neutral water ph = poh at any temp. 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 These are very important? Make sure you study these! At any temp: ph + poh = pkw At 25 0 C: ph + poh =