Chapter 3: Acid Base Equilibria. HCl + KOH KCl + H 2 O acid + base salt + water

Transcription

1 Chapter 3: Acid Base Equilibria HCl + KOH KCl + H 2 O acid + base salt + water

2 What is an acid? The Arrhenius concept proposed that acids are substances that produce hydrogen ions (H + ) in aqueous solutions. If this: HA (aq) + H 2 O (l) H 3 O (aq) + A - (aq) or the abbreviated equal: HA (aq) H + + A - is the general form of the acid reaction, then we can calculate an equilibrium constant (K a ) for the reaction. K a is called the acid dissociation constant. K a = [H + ][A - ] / [HA]. A strong acid is one that undergoes significant dissociation and has a very large K a. A weak acid only partially dissociates and has a relatively small K a. pk a = -logk a

3 Table 3-1 Dissociation Constants for Acids at 25 C Acid Formula pk a1 pk a2 pk a3 Hydrochloric HCl ~-3 Sulfuric H 2 SO 4 ~ Nitric HNO 3 0 Oxalic H 2 C 2 O Phosphoric H 3 PO Hydrofluoric HF 3.18 Formic HCOOH 3.75 Acetic CH 3 COOH 4.76 Carbonic H 2 CO Hydrosulfuric H 2 S 7.03 >14 Boric H 3 BO >14 Silicic H 4 SiO

4 As you noticed from the previous slide (table 3-1), acids can contain more than one acidic proton. i.e. H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Polyprotic acids will have more than one pk a Remember: pk a = -logk a Acids that undergo significant dissociation have a negative pk a, and acids that only partially dissociate have a positive pk a.

5 Table 3-1 Dissociation Constants for Acids at 25 C Acid Formula pk a1 pk a2 pk a3 Hydrochloric HCl ~-3 Sulfuric H 2 SO 4 ~ Nitric HNO 3 0 Oxalic H 2 C 2 O Phosphoric H 3 PO Hydrofluoric HF 3.18 Formic HCOOH 3.75 Acetic CH 3 COOH 4.76 Carbonic H 2 CO Hydrosulfuric H 2 S 7.03 >14 Boric H 3 BO >14 Silicic H 4 SiO

6 What is a base? The Arrhenius concept proposed that a base is a substance that produces OH - ions in aqueous solution. If this: B (aq) + H 2 O (l) BH + (aq) + OH - (aq) is the general form of the base reaction, then we can calculate an equilibrium constant (K b ) for the reaction. K b is called the base dissociation constant. K b = [BH + ][OH - ] / [B]. A strong base is one that undergoes essentially complete dissociation and has a large K b. A weak base only partially dissociates and has a relatively small K b. pk b = -logk b

7 Table 3-2 Dissociation Constants for Bases at 25 C Base (Hydroxide) Formula pk b1 pk b2 pk b3 Methylamine CH 3 NH Ammonium NH 4 (OH) 4.7 Magnesium Mg(OH) Pyridine C 5 H 5 N 8.8 Manganese Mn(OH) Ferrous Fe(OH) Al, amorphous Al(OH) Al, gibbsite Al(OH) Ferric, amorphous Fe(OH) *As it was with acids, base may also require more than one step to complete dissociation. You may also notice the absence of NaOH. For our purposes, you may assume this base completely dissociates.

8 The dissociation of water and ph H 2 O H + + OH - and K w = [H + ][OH - ] / [H 2 O] = [H + ][OH - ] Where K w is the equilibrium constant for water. K w varies as a function of temperature. (See table 3-3.) K w(25 C) = = [H + ][OH - ] *Remember that K w is a constant, therefore if you know the K w and the concentration of either H + or OH -, you can find the remaining unknown concentration.

9 For example: If [H + ] for a solution is 10-3, what is [OH-] in that solution? [OH - ] = The ph scale was designed to simplify the description of the acidity of a solution. The ph is often called the hydrogen ion exponent. ph = -log[h + ] Therefore, if, for example [H + ] = 10-3, then the ph = 3 Pure water at 25 C has a ph of 7. This means that the number of H + ions and the number of OH - ions are equal at 10-7 a piece.

10 Why is knowing the specific [H+] important?

11 ph of Natural Waters

12 Pre-industrial and current samples of Globigerina bulloides from same location. Latter shows extensive erosion with a ten-fold increase in holes.

13 Ocean surface ph - historical values and projected future values based on current emission projections. As a comparison, ph levels during the late Permian (the time of the great dying) were ~ lower that normal.

14 Corals are being hit on two fronts, bleaching from temperature rise and dissolution from a decreasing ph.

15 Back to the calculations

16 Determine the ph of a solution with a [OH - ] of 3 X Recall: K w(25 C) = = [H + ][OH - ] / 3 X 10-5 = [H + ] [H + ] = 3.33 X ph = -log(3.33 X 10-9 ) = 8.48 Is this solution acidic or basic?

17 ph of river systems?

18 Not so natural water.acid mine drainage.

19 It is very important to be familiar with the carbonatecarbonic acid system. It is one of the most important systems in the environment.

20 You might want to write these equations down. You will, without a doubt, see them again. CO 2 + H 2 O H 2 CO 3 H 2 CO 3 H + + HCO 3 - HCO 3 - H + + CO 3 2- These describe the formation and dissociation of carbonic acid. Theoretically, what would happen to the concentration of CO 2 if more H + were added to the system? We will address this later.

21 CO 2(g) + H 2 O H 2 CO 3(aq) where [H 2 CO 3(aq) ] = K CO2 P CO2 and K a1 = [H + ][HCO 3- ] / K CO2 P CO2 The dissociation of H 2 CO 3 is a 2-step process! H 2 CO 3(aq) H + + HCO - 3 HCO - 3 H + + CO 2-3 K a1 = [H + ][HCO 3- ] / [H 2 CO 3(aq) ] and K a2 = [H + ][CO 2-3 ] / [HCO 3- ] Further at 25 C: [H 2 CO 3(aq) ] / [HCO 3- ] = [H + ] / K a1 = [H + ] / And [HCO 3- ] / [CO 2-3 ] = [H + ] / K a2 = [H + ] /

22 CO 2 + H 2 O CH 2 O + O 2 You should all be familiar with the basic equation for photosynthesis. How is it similar to the carbonic acid equation? More accurate equation for photosynthesis: 6CO 2 + 6H 2 O C 6 H 12 O 6 + 6O 2 What does this imply?

23 You can consider respiration to be the opposite of photosynthesis using the simplest of terms CH 2 O + O 2 CO 2 + H 2 O

24 Given: [H 2 CO 3(aq) ] / [HCO 3- ] = [H + ] / K a1 = [H + ] / [HCO 3- ] / [CO 3 2- ] = [H + ] / K a2 = [H + ] / How would you expect H 2 CO 3 to affect the ph of natural waters? It depends upon the level of dissociation.

25 How does the weathering of calcium, silica and carbonate species change the atmospheric concentration of carbon dioxide? Images from midfind.com

26 CaSiO 3 + 3H 2 O + 2CO 2 Ca (HCO 3 - ) + H 4 SiO 4 wollastonite + water + carbon dioxide calcium + bicarbonate + silicic acid From pore spaces &/or atmosphere and: Ca (HCO 3 - ) CaCO 3 + H 2 O + CO 2 and: CaCO 3 + SiO 2 CaSiO 3 + CO 2 calcite + quartz wollastonite + carbon dioxide

27 What other factors can change the ph of natural waters? Table 3-5. Examples of processes that control the CO 2 content & ph of surface and ground waters Process Reaction ph Temperature change Increase T, decrease solubility of CO 2 (g) Increases Decrease T, increase solubility of CO 2 (g) Decreases Photosynthesis 6CO 2(g) + 6H 2 O C 6 H 12 O 6 + 6O 2(g) Increases Respiration C 6 H 12 O 6 + 6O 2 (g) 6CO 2 (g) + 6H 2 O Decreases Anaerobic decay 2CH 2 O CH 4 (g) + CO 2 (g) Decreases Denitrification 5CH 2 O + 4NO H + 5CO 2 (g) + 2N 2 (g) + 7H 2 O Increases Dissolution of carbonate CaCO 3 calcite + 2H + Ca 2+ + H 2 O + CO 2 (g) Increases Precipitation of carbonate Ca 2+ + H 2 O + CO 2 (g) CaCO 3 calcite + 2H + Decreases Weathering of Al-silicate minerals 2KAlSi 3 O 8 feldspar + 2CO 2 (g) + 11H 2 O Increases Al 2 Si 2 O 5 (OH) 4 kaolinite + 2K + + 2HCO - + 4H 4 SiO 4 (aq)

28 Lets consider the effects of water in contact with atmospheric carbon dioxide. CO 2(aq) + H 2 O (l) H 2 CO 3(aq) Given this relationship between CO 2 in solution and water, would you expect the surface waters of the ocean to be more or less acidic than the ocean waters out of contact with the atmosphere? In general, surface waters are more acidic at the surface and in coastal waters (for another reason ) With the CO 2 H 2 O system there are two end-member cases: An open system: in equilibrium with atmospheric CO 2. A closed system: isolated from atmospheric CO 2.

29 Let s recall problem 25 from the problem set 2. What did we discover about the ionic charges in a solution? The total positive (cation) charge and total negative (anion) charge in a solution must be equal. For example: for the CO 2 H 2 O system, the charge balance equation is written: m H + = m HCO m CO m OH - Where m is the molar concentration of each species in solution.

30 The Carbonic acid Carbonate System is regulated by not only the presence of H 2 CO 3, but also the presence of the minerals calcite or aragonite (CaCO 3 ). CaCO 3calcite Ca 2+ +CO 2-3 K sp = [Ca 2+ ][CO 2-3 ] Final charge balance equation becomes: m H + + 2m Ca 2+ = m HCO m CO m OH -

31 Now let s do some problems from the book

32 Example 3-3 (pg. 66) Calculate the ph of rainwater in equilibrium with atmospheric CO 2. First of all, this book was first published in 2004, thus the data is more than 10 years old. What are the atmospheric CO 2 levels today? For the present day atmosphere = (400ppm) P CO 2 = atm. [H 2 CO 3 ] = K CO 2 P CO2 = ( )( ) = K a =[H + ][HCO 3- ] / [H 2 CO 3 ] [H + ][HCO 3- ] = K a [H 2 CO 3 ] = ( )( ) = [ ] If you assume that [H + ] = [HCO 3- ] then [H + ] = ( ) 0.5 = this means that the ph of acid rain is less than It is important to note that the book suggest acid rain has a ph of What does this imply if the atmospheric CO 2 concentrations approach 500ppm? Remember this is a logarithmic scale.. The ph would be 5.56.

33 Example 3-4 A groundwater sample has a measured ph of 6.84 and HCO 3 - of 460mg L -1. We will assume that activity equals concentration. At 25⁰C, calculate the P CO 2 for this groundwater sample. We must first convert the measured concentration of HCO 3 - to moles per liter. The atomic weight of HCO 3 - = 61.0g [HCO 3 - ] = 460 x 10-3 g L -1 / 61g/mol = 7.54 x 10-3 mol L -1 The using equation 3-15: logp CO 2 = -ph + log([hco 3 - ]/K a 1 K CO2 ) logp CO 2 = log([7.54 x 10-3 ] / ) = P CO 2 = How does this compare with atmospheric concentrations? What does this imply about the groundwater? Is it close to a pollution source, anaerobic or aerobic decay or in supersaturated in CaCO 3?

34 Example 3-5 Calculate the concentration of each carbonate species in a solution at 25⁰C when C T = 1 x 10-3 mol L -1 and ph = 5.7. Which species do you expect to be dominate? α H = 1 + K a 1 /[H+ ] + K a 1 K a2 /[H+ ] 2 = 1 + ( )/( ) + ( )( )/( ) 2 = [H 2 CO 3 ] = C T /α H = 1 x 10-3 mol L -1 /1.224 = 8.17 x 10-4 mol L -1 [HCO 3- ] = C T K a 1 /[H+]α H = (1 x 10-3 )( )/( )(1.224) = 1.83 x 10-4 mol L -1 [CO 3 2- ] = C T K a1 K a2 /[H + ] 2 α H = (1 x 10-3 )( )( )/( ) 2 (1.224) = 4.29 x 10-9 mol L -1 Do these concentrations make sense given a ph of 5.7

35 ph = 5.7: [H 2 CO 3 ] = 8.17 x 10-4, [HCO 3- ] = 1.83 x 10-4, [CO 3 2- ] = 4.29 x 10-9

36 Acidity and Alkalinity Acidity is the capacity of water to donate protons. Also described as the ability of a solution to neutralize bases. Alkalinity is the capacity of water to accept protons. Also described as the ability of a solution to neutralize acids. Nonconservative species: species whose abundances vary as a function of ph or some other intensive variable (i.e.p & T). Conservative species: species whose abundances do not vary as a function of ph or some other intensive variable (i.e.p & T).

37 What is the easiest way to determine the concentration of either H + or OH -? Titration. C t X V t = C s X V s Where C t is the concentration of the titrant, V t is the volume of the titrant, C s is the concentration (acidity or alkalinity) of the unknown solution and V s is the volume of the unknown solution.

38 Example mL of an acidic solution is titrated with a 100 meq L -1 NaOH solution. Neutrality (ph = 7) is achieved after 50mL of titrant have been added to the acid solution. Calculate the acidity of the solution. C s = (C t x V t ) / V s = ((100meq L -1 )(50 x 10-3 L)) /100 x 10-3 L = 50 meq L -1 = total acidity of the unknown.

39 Salts 1. strong acid with a strong base. makes a neutral solution. 2. weak acid with a strong base. makes a basic solution. 3. weak base with a strong acid. makes an acidic solution. 4. weak acid with a weak base. makes either a basic or an acidic solution depending on the relative strength of the ions. **Most, but not all, minerals can be considered to be salts of weak acids and strong bases.

40 In every family there s always an oddball. Amphoteric Hydroxides: hydroxides that can behave as either and acid or a base. This behavior varies as a function of ph. For example: Al(OH) 3 behaves as a base in an acidic solution and forms aluminum salts. In a basic solution, it behaves as an acid H 3 AlO 3 and forms salts with AlO K A is the equilibrium constant for an amphoteric reaction. A table of these constants is listed on page 75 of the text.

41 Buffers A buffered solution is a solution that resists changes in ph when either hydrogen or hydroxyl ions are added to the solution. A buffer is a weak acid and its salt or a weak base and its salt. Let s go back to our buddy LeChâtelier s principle and look at the following reactions. NaC 2 H 3 O 2 Na + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O 2 - What would happen if a strong acid were added to a solution containing both NaC 2 H 3 O 2 and HC 2 H 3 O 2?

42 NaC 2 H 3 O 2 Na + + C 2 H 3 O 2 - HC 2 H 3 O 2 H + + C 2 H 3 O 2 - HCl H + + Cl - When you add additional H+ to this buffered solution, what happens to the concentration of: HC 2 H 3 O 2? C 2 H 3 O 2 -? NaC 2 H 3 O 2? Increases Decreases Decreases What ultimately happens to the H + added to the solution? It combines with C 2 H 3 O 2 - to make HC 2 H 3 O 2 until all the is NaC 2 H 3 O 2 used up.

43 Example 3-13 What would happen to the ph of a 1 liter solution, containing carbonic acid, if 10-4 mol of H + ions are added to the solution? At ph = 7, T = 25 C, [HCO 3- ] = 10-3 mol/l, pk a1 = 6.35, therefore, [H 2 CO 3 ] = mol/l H + + HCO 3 - H 2 CO 3(aq) According to LeChâtelier s principle the concentration of will HCO 3 - decrease by 10-4 mol/l, and the concentration of H 2 CO 3(aq) will increase by 10-4 mol/l. ph = pk a1 + log([hco 3- ] / [H 2 CO 3 ]) = log([ ] / [ ]) = 6.79 If the ph of the solution would have been 4 without the H 2 CO 3 ; and with the H 2 CO 3 the ph is 6.79, then it can be said that the solution was successfully buffered.

44 The previous exercise demonstrated the Henderson- Hasselbalch equation. Consider the generic: H+ + A- HA. ph = -logka + log ([A-] / [HA]) Remember that buffers are important in the natural environment because they control the impact of acid or base additions on natural waters. This is an acidic lake in Norway. Norway s geology consists primarily of crystallines and metamorphites and little limestone. The acidity was caused by acid rain.

45 Buffering Capacities of Natural Waters Water is an effective buffer only at very high or very low ph. Buffering capacity: a measure of the amount of H + or OH - ions a solution can absorb without significant change in ph.

46 Buffering Capacity of H 2 CO 3 Buffering capacity of CaCO 3 H 2 CO 3 system

47 Figure 3-9 summarizes the buffering capacities of waters in contact with minerals Does the fate of the Norwegian acid lake seem clearer?

48 Exercise 3-15 takes a look at the buffering index for the calcite-carbonic acid system. Calculate the buggering index for the system calcite-carbonic acid at ph = 6, C T = 1 x 10-2, and T = 25⁰C. Using equation After several steps and plugging and chugging B = 2.34 x x = eq L -1 ph -1 = meq L -1 ph -1

49 Before we leave the discussion on acids and bases, we should also mention the importance of silicic acid (H 4 SiO 4 ). This important acid is a produced during the weathering of silicate minerals the most abundant mineral group in the crust of the earth. We need only consider the first two dissociation stops with this tetraprotic acid. spx?main=7&questionid=4438&expan dtype=1 At ph = 9.83, [H 4 SiO 4 ] = [H 3 SiO 4- ]. This implies that for ph < 9.83, H 4 SiO 4 is the dominant species. Only at ph > rarely if ever found in nature is H 2 SiO 4 2- the dominant species. 2KAlSi 3 O 8 + 9H 2 O + 2H+ Al 2 Si 2 O 5 (OH) 4 + 2K + + 4H 4 SiO 4

50 Problem Set 3 Chapter 3 #s: 3, 8, 14, 19, 21, 24, 27, 35, 36 These are due March 7 th.