1 Chapter 10 Monoprotic Acid and Bases

2 Strong Acids and Bases Compound that when dissolved in water will fully dissociate. This is a factor of our very universal solvent and its’ special properties. This is called the leveling effect of water. Our strong acids are HCl HBr HI HNO 3 HClO 4 H 2 SO 4 (first ionization)

3 Strong Acids The acid dissociation equilibrium HA + H 2 O = H 3 O + + A - The equilibrium constant values for our strong acids: AcidK a HCl7900 HBr HI HNO 3 25

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6 Strong Acids The strongest of the common strong acids is perchloric acid. (HClO 4 ) This strengths can be shown be going to other solvents other than water. Acidic acid is a common choice.

7 Strong Acid and Bases Our strong bases are NaOH KOH LiOH Sr(OH) 2

8 Determining pH Since they completely dissociate we get the same concentration of H + as the dissolved acid or base. So if a solution is made up to be 1.00x10 -2 M HCl then the pH will be pH = - log [H+] = -log (1.00x10 -2 ) = 2.?????? (How many sig figs?) More correctly we should correct for activity so we should have pH = -logA H+ = -log  [H+] = -log (0.914)(1.00x10 -2 ) = (How many sig figs?) But we will not bother to correct unless called on to do so.

9 Strong Acids pH of strong base solutions. pH of a 1.0x10 -3 M NaOH solution. [Na+] =.0010 and [OH-] = Then pOH is 3.00 Since [H + ][OH - ] = K w pH + pOH = Then pH of this solution is K w varies with temperature so you need to account for that also.

10 Strong Acids What if we make up a solution that is 1.0x10 -5 M HNO 3. The pH would be If we were to dilute this solution by 1000 fold. That is take 1.00 mL and dilute to 1.00 Liters. What is the pH then.

11 Strong Acids It can not be pH 8. That would mean that we have made up a solution that is basic from and acid and water??? How would you calculate this? This is a case for Chapter 9. Systematic treatment of equilibrium.

12 Strong Acids What do we know? C HNO3 = 1.00x10 -8 From this we can see that C NO3- = 1.00x10 -8 What is the charge balance of this sytem. [H + ] = [NO 3 - ] + [OH - ] We also have our K w expression. So [H + ] = C NO3 + K w /[H + ]

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14 Weak Acids/Bases These are acids and bases that do not fully dissociate when placed in aqueous solution. Examples Acetic Acid Benzoic Acid Hydrofluoric Acid Ammonia

15 Weak Acids/Bases Weak acid equilibrium HA = H + + A -

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17 Weak Acids and Bases Weak Base equlibrium (Also called hydrolysis constant) B + H 2 O = BH + + OH -

18 Conjugate Pairs Formic AcidFormate Ion An acid / conjugate base pair

19 Conjugates Ammonia Ammonium Ion BaseConjugate Acid

20 Table G The Table in the back of the book. Starting on page AP12 lists all acids and bases as acids. It will give the proper base name but show the structure for the conjugate acid form. When there is more than one acidic group then the book will indicate which proton goes with the given pKa.

21 Table G 8- Hydroxyquinoline 4.91 (NH) 9.81 (OH)

22 Buffers When you add a weak acid and its’ conjugate then you will get a buffer. Buffers resist the rapid change in pH when acid or base is added to the solution. pH = pKa + log (base/acid) pH control is important since many processes are pH dependent.

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25 Buffers Made two different ways Mix an acid and its conjugate base Acetic acid and sodium acetate Ammonium Chloride and aqueous ammonium Prepare a solution of an acid or base and generate the conjugate by addition of strong acid or base. Acetic acid and add NaOH

26 Buffers When adding both forms the concentration of the buffer will be the sum of the concentration of both forms When generating from strong conjugate then add the moles of acid needed at the final volume.

27 Preparation Weight out the number of moles of buffer needed and dilute to ~80% final volume. Place pH electrode into solution Add strong acid/base until pH is reached. Dilute to mark

28 Buffer Capacity The amount of acid or base that can be added before the buffer is consumed.  = dC b / dpH = - dCa / dpH

29 Buffer pH Ionic Strength Temperature

30 What about extreme conditions F HA + F A- = [HA] + [A - ] [Na + ] + [H + ] = [OH - ] + [A - ] [HA] = F HA – [H + ] + [OH - ] [A-] = F A- + [H + ] – [OH - ] pH = pKa + log {corr A- / corr HA}

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