First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, 11-13

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1 Announcements First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, Please keep up with the work (lots of problems in this Chapter) and see me if you have questions.

2 Ion-product of Water is Constant = H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Acid Base Acid Base If we measure pure water equilibrium concentrations we find: [H3O + ] = [OH - ] = 1 X 10-7 We can define an equilibrium constant for water, Kw as: K w = [H 3O + ][OH - ] [H 2 O] 2 = [H3O + ][OH - ] = 1 X In any dilute aqueous solution the ionproduct constant of water, Kw, = 1.0 X 10-14

3 Implications of K w = [H3O + ][OH - ] = 1 X [H3O + ] [OH - ] Kw 1 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X 10-14

4 [OH - ] and [H + ] vary over a huge range If we plot a few points of [OH - ] vs [H + ] then we quickly realize we can not plot over the various powers of 10 that can exist in solution. In 1909, Sorenson suggests using a logarithmic plot and this makes things easy to see and to plot. It defines ph!

5 Logarithmic Plot of poh vs ph By plotting----- log [OH] vs - log [H] we can now see 14 orders of magnitude (14 powers of 10) in a single graph! The -log[h 3 O + ] is called the ph.

6 ph = -log[h3o + ] ph = -log [H + ] = -log [H3O + ] p = -log hydrogen ion or hydronium ion concentration at equilibrium Danish biochemist Sorenson proposed using logarithmic mathematics to condense the range of H3O + and OH - concentrations to a more convenient scale. Each ph unit or number represents a power of 10. A solution with a ph of 4 is 10X more than one with ph=3, 100X more than ph=2 and 1000X more than ph = 1.

7 A Logarithm is Just An Exponent Let y be any number that your heart desires. Let s also let b be any number we wish and let s call it the base. Lastly let s call the power we raise the base to an exponent and label it as x y = b x 100 = 10 x Let s make the base, b = 10 and y =100. The logarithm of 100, or in shorthand log (100) is the value of the exponent, x needed to raise a base to in order to obtain that number. The value of the exponent log (100) = x = 2 required in the equation above!

8 What is a logarithm? What is the log10 of the following numbers? 1000, 100, 10, 1, 0, 0.1, 0.01, log = log 1000 = x log = log 100 = x log 10 = x log 1 = x log 0 = x log 0.1 = x log 0.01 = x Determining the log 10 is the same as computing x in this equation! 10 X = X = X = X = 1 10 X = 0 10 X = X = 0.01

9 log and ln are common logarithms In math we use two common logarithm functions. They differ by their base only! log and ln log 10 and log e = log = ln

10 Properties of Logs log (A B) = log A + log B log ( ) A B = log A log B log A n = n log A Rules for significant figures: Only count to the right of the decimal point. Log 23.5 = Significant Figures

11 Antilogarithms is the inverse Sometimes we know the base and the exponent, but we need to compute the number. We can write it like this: log (?) = 2 In words we would say: The log10 of what number is equal to 2? Finding this number is called taking the antilogarithm Antilog (log (?)) = Antilog (2)? = 10 2 = 100 The antilog10 is the same as the base 10!

12 Log and Exponential Forms Logarithmic Form ph = -log [H 3 O + ] Exponential Form 10 -ph = [H 3 O + ] poh = -log [OH - ] 10 -poh = [OH - ] pk a = -log K a 10 -pka = pk a When H 3 O +, OH -, K a are large values----the pk s are small values. Strong acids have large H 3 O and K a therefore small ph and small pk a

13 [H3O + ] ph [OH - ] poh Relation ships between [H 3 O + ], ph, [OH - ] and poh

14 Kw is related to ph via logarithm Define some terms: ph = -log[h 3 O + ] poh = -log[oh - ] pkw = -log Kw K W = [H 3 O + ][OH - ]= 1.0 X Now take the -log of both sides of the ion-water product constant -log K W = - log([h 3 O + ][OH - ]) = -log (1.0 X ) -log K W = - log[h 3 O + ] + -log[oh - ] = -log (10-14 ) pk W = ph + poh = -(-14) pk W = ph + poh = 14

15 Connecting The Dots of Kw and pkw K W = [H3O + ] [OH] = 1 x pk W = ph + poh = 14 Remember these they will be helpful from this day forward It s very very easy to measure ph in the laboratory!

16 Representative Acids, Ka and pka Acid Ionization Equilibrium Ka pka Hydroiodic Acid HI + H2O <----> H3O + + I - 3 X Hydrochloric Acid HCl + H2O <----> H3O + + Cl X Sulfuric Acid H2SO4 + H2O <----> H3O + + HSO Hydronium Ion H3O + H2O <----> H3O + + IO Nitric Acid HNO3 + H2O <----> H3O + + NO Iodic acid HIO3 + H2O <----> H3O + + IO X Chlorous Acid HClO2 + H2O <----> H3O + + ClO2-1.1 X Phosphoric Acid H3PO4 + H2O <----> H3O + + H2PO4-7.6 X Nitrous Acid HNO2 + H2O <----> H3O + + NO2-7.2 X Hydrofluoric Acid HF + H2O <----> H3O + + F X Formic Acid HCOOH + H2O <----> H3O + + HCO2-1.8 X Acetic Acid CH3CO2H + H2O <----> H3O + + CH3CO2-1.8 X Hydrosulfuric Acid H2S + H2O <----> H3O + + HS X Hypochlorous Acid HClO + H2O <----> H3O + + ClO X Hydrocyanic Acid HCN + H2O <----> H3O + + CN X

17 Ways to Describe Acid Strength Property Acid Base Ka Ka is large, pka is small Ka is small, pka is large Position of Ka Far to the right Far to the left Strength of Conjugate Base vs water A - much weaker than water A - much stronger than water

18 Acid-Bases Equilibrium Problems HCl(aq) + H 2 O H 3 O + (aq) + Cl (aq) 0.004% at equilibrium % at equilibrium STRONG ACID CH 3 CO 2 H(aq) + H 2 O H 3 O + (aq) + CH 3 CO 2 (aq) 98.7% at equilibrium 1.3% at equilibrium WEAK ACID

19 Strong Acid = Strong electrolytes Strong Acids 1 No HA Weak Acids 2 Not dissociated at all 3

20 Common Strong Acids and Bases These are 100% ionized so that [HCl] = [H + ] = [H3O + ] for acids and [NaOH] = [OH - ] for bases! These common strong acids and bases fit the Arrhenius definition of acid and base. Acids donate H + while bases donate OH -

21 Acid-Base Equilibrium HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) K a = [H 3O + ][A - ] [HA] Defining Ka the acid dissociation constant at equilibrium B + H 2 O(l) BH + (aq) + OH - (aq) K b = [BH+ ][OH - ] [B] The magnitude of the equilibrium constant called Kb tells us how strong a base is!

22 Sample Problem What is the ph of a 2 x 10-3 M HNO 3 solution? HNO 3 is a strong acid it is100% dissociated. Initial Change Equilibrium HNO 3 (aq) + H 2 O (l) M M 0.0 M H 3 O + (aq) + NO - 3 (aq) 0.0 M 0.0 M M M M M ph = -log [H 3 O + ] = -log(0.002) = 2.7

23 What is the ph of a 1.8 x 10-2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base 100% dissociation. Initial Change Equilibrium Ba(OH) 2 (s) M M 0.0 M Ba 2+ (aq) + 2OH - (aq) 0.0 M 0.0 M M M M M pkw = ph + poh = ph = poh ph = (- log(0.036)) = 12.56

24 Sample Problem The ph of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H + ion concentration of the rainwater? ph = -log [H + ] antilog(-ph) = antilog (log [H + ]) [H + ] = 10 -ph = = 1.5 x 10-5 M The OH - ion concentration of a blood sample is 2.5 x 10-7 M. What is the ph of the blood? ph + poh = poh = -log [OH - ] = -log (2.5 x 10-7 ) = 6.60 ph = poh = = 7.40

25 Problem: Calculating [H 3 O + ], ph, [OH - ], and poh In a restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and M HNO 3. Calculate [H 3 O + ], ph, [OH - ], and poh of the three solutions at 25 o C.

26 In a restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and M HNO 3. Calculate [H 3 O + ], ph, [OH - ], and poh of the three solutions at 25 o C. SOLUTION: For 2.0 M HNO 3, [H 3 O + ] = 2.0 M, and ph = -log [H 3 O + ] = = ph [OH - ] = K w / [H 3 O + ] = 1.0 x /2.0 = 5.0 x M; poh = For 0.3 M HNO 3, [H 3 O + ] = 0.30 M and -log [H 3 O + ] = 0.52 = ph [OH - ] = K w /[H 3 O + ] = 1.0 x /0.30 = 3.3 x M; poh = For M HNO 3, [H 3 O + ] = M and -log [H 3 O + ] = 2.20 = ph [OH - ] = K w / [H 3 O + ] = 1.0 x /6.3 x 10-3 = 1.6 x M; poh = 11.80

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