ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA Acids- taste sour Bases(alkali)- taste bitter and feel slippery Arrhenius concept- acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions Bronsted-Lowry model- acids are proton (H ) donors and bases are proton acceptors Lewis model- acids are electron pair acceptors while bases are electron pair donors (not tested on AP test) hydronium ion (H 3 O )- formed on reaction of a proton with a water molecule. H and H 3 O are used interchangeably in most situations. HA(aq) H 2 O(l) H 3 O (aq) A (aq) Acid Base Conjugate Conjugate Acid Base conjugate base- everything that remains of the acid molecule after a proton is lost conjugate acid- base plus a proton Acid dissociation constant (K a ) K a = [H 3 O ][A ] or K a = [H ][A ] [HA] [HA] Strong acid - mostly dissociated (essentially 100%) - equilibrium lies very far to the right - a strong acid yields a weak conjugate base (much weaker than H 2 O) Weak acid- mostly undissociated - equilibrium lies far to the left - has a strong conjugate base (stronger than water) The weaker the acid, the stronger its conjugate base (inverse relationship). Sketch what five molecules of HF (weak) and five molecules of HCl (strong) would look like in a beaker of water. Common strong acids -all aqueous solutions (Know these!) H 2 SO 4 (sulfuric) HCl (hydrochloric) HNO 3 (nitric) HClO 4 (perchloric) HI (hydroiodic) HBr (hydrobromic) HF HCl Acid/Base/Salt Equilibria Notes pg.1 Kristen Jones 6/21/2013

Sulfuric acid is a diprotic acid which means that it has two acidic protons. The first (H 2 SO 4 ) is strong and the second (HSO 4 ) is weak. Oxyacids- most acids are oxyacids (contain oxygen) - the acidic proton is attached to O Weak oxyacid examples: H 3 PO 4 (phosphoric) HNO 2 (nitrous) HOCl (hypochlorous) Within a series, acid strength increases with increasing numbers of oxygen atoms. For example: HClO 4 > HClO 3 > HClO 2 > HClO and H 2 SO 4 > H 2 SO 3 (Electronegative O draws electrons away from O-H bond) Acid strength increases with increasing electronegativity of oxyacids. For example: HOCl>HOBr>HOI>HOCH 3 Organic acids- have carboxyl group - usually weak acids CH 3 COOH acetic acid C 6 H 5 COOH benzoic acid Hydrohalic acids- H is attached to a halogen (HCl, HI, HBr, HF) HF is the only weak hydrohalic acid. Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely dissociate. Weak acid strength is compared by the K a values of the acids. The smaller the K a, the weaker the acid. Strong acids do not have K a values because the [HA] is so small and cannot be measured accurately. Amphoteric substance- Substance that can act as an acid or as a base. Ex. H 2 O, NH 3 -To be amphoteric, a substance must be able to both donate and accept a proton. Autoionization of water H 2 O H 2 O H 3 O OH base acid conjugate conjugate acid base Ion product constant for water (K w ) K w = [H 3 O ][OH ] K w = [H ][OH ] At 25 o C, K w = 1 10 14 because [H ] = [OH ] = 1 10 7 M No matter what an aqueous solution contains, at 25 o C [H ][OH ] = 1 10 14 Neutral solution [H ] = [OH ] Acidic solution [H ] > [OH ] Basic solution [H ] < [OH ] K w varies with temperature ph = log [H ] If [H ] = 1.0 10 7 M, ph = 7.00 Acid/Base/Salt Equilibria Notes pg.2 Kristen Jones 6/21/2013

Significant figures in ph and other log values: The number of decimal places in the log value should equal the number of significant digits in the original number (concentration). poh = log [OH ] pk = log K ph and poh are logarithmic functions. The ph changes by 1 for every power of 10 change in [H ]. ph decreases as [H ] increases. ph poh = 14 [H ] = antilog(ph) or [H ] = 10 ph [OH ] = antilog(poh) or [OH ] = 10 poh Calculating ph of Strong Acid Solutions Calculating ph of strong acid solutions is generally very simple. The ph is simply calculated by taking the negative logarithm of the concentration of a monoprotic strong acid. For example, the ph of 0.1 M HCl is 1.0. However, if the acid concentration is less than 1.0 10 7, the water becomes the important source of [H ] and the ph is 7.00. The ph of an acidic solution can not be greater than 7 at 25 o C!!!!! Another exception is calculating the ph of a H 2 SO 4 solution that is more dilute than 1.0 M. At this concentration, the [H ] of the HSO 4 must also be calculated. Ex. Calculate the [H ] and ph in a 1.0 M solution of HCl. HCl is a strong monoprotic acid, therefore its concentration is equal to the hydrogen ion concentration. [H ] = 1.0 M ph = log (1.0) = 0.00 Ex. Calculate the ph of 1.0 10 10 M HCl. Since the [H ] is less than 1.0 10 7, the [H ] from the acid is negligible and the ph = 7.00 Calculating ph of Weak Acid Solutions Calculating ph of weak acids involves setting up an equilibrium expression. Always start by writing the equation, setting up the acid equilibrium expression (K a ), defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the K a expression and solving for x. Ex. Calculate the ph of a 0.100 M solution of acetic acid. The K a of acetic acid is 1.8 10 5. HC 2 H 3 O 2 H C 2 H 3 O 2 K a = [H ][C 2 H 3 O 2 ] = 1.8 10 5 [HC 2 H 3 O 2 ] Reaction HC 2 H 3 O 2 H C 2 H 3 O 2 Initial 0.100 0 0 Change x x x Equilibrium 0.100 x x x Often, the x in a K a expression can be treated as negligible. 1.8 10 5 = (x)(x) x 2 x = 1.34 10 3 0.100 x 0.100 1.34 10-3 is less than 5% of 0.100 (it is 1.34%), so our assumption is valid. Acid/Base/Salt Equilibria Notes pg.3 Kristen Jones 6/21/2013

ph = log [H ] [H ] = x = 1.34 10 3 ph = log 1.34 10 3 ph = 2.87 Quadratic Equation Problem Ex. Calculate the ph of a 1.00 10-4 M solution of acetic acid. The K a of acetic acid is 1.8 10 5. HC 2 H 3 O 2 H C 2 H 3 O 2 K a = [H ][C 2 H 3 O 2 ] = 1.8 10-5 [HC 2 H 3 O 2 ] Reaction HC 2 H 3 O 2 H C 2 H 3 O 2 Initial 1.00 10-4 0 0 Change x x x Equilibrium 1.00 10-4 x x x 1.8 10 5 = (x)(x) x 2 x = 4.2 10 5 1.00 10 4 x 1.00 10-4 4.2 10-5 is greater than 5% of 1.00 10 4. This means that the assumption that x was negligible is invalid and x must be solved for using the quadratic equation. On the AP test, the quadratic equation will not need to be used. Use of the quadratic equation: x 2 1.8 10-5 x - 1.8 10-9 = 0 x = -1.8 10-5 ± (1.8 10-5 ) 2 4(1)( 1.810-9 ) 2(1) x = 3.5 10-5 and 5.2 10-5 Since a concentration can not be negative, x= 3.5 10-5 M x = [H ] = 3.5 10-5 ph = log 3.5 10-5 = 4.46 Calculating ph of polyprotic acids All polyprotic acids dissociate stepwise. Each dissociation has its own K a value. As each H is removed, the remaining acid gets weaker and therefore has a smaller K a. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton. Except for H 2 SO 4, polyprotic acids have K a2 and K a3 values so much weaker than their K a1 value that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H ] obtained from this 2nd and 3rd dissociation is negligible compared to the [H ] from the 1st dissociation. Because H 2 SO 4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem (and it won t be asked on the AP test). Ex. Calculate the ph of a 1.00 10 2 M H 2 SO 4 solution. The K a of HSO 4 is 1.2 10 2. Because H 2 SO 4 is a strong acid in its first dissociation, we start with stoichiometry and a BCA (Before, Change, After) chart. H 2 SO 4 H HSO 4 Before 1 10 2 0 0 Change 1 10 2 1 10 2 1 10 2 After 0 1 10 2 1 10 2 Acid/Base/Salt Equilibria Notes pg.4 Kristen Jones 6/21/2013

The products are then brought down to the equilibrium chart (RICE). R HSO 4 H 2 SO 4 I 1 10 2 1 10 2 0 C x x x E 1 10 2 x 1 10 2 x x 1.2 10 2 = [H ][SO 4 2 ] [HSO 4 ] 1.2 10 2 = (1 10-2 x)(x) (1 10-2 x) Using the quadratic equation, x = 4.52 10 3 [H ]= 1 10 2 (4.52 10 3 ) = 1.45 10 2 ph = 1.84 Determination of the ph of a Mixture of Weak Acids Only the acid with the largest K a value will contribute an appreciable [H ]. Determine the ph based on this acid and ignore any others. Determination of the Percent Dissociation (or Ionization) of a Weak Acid Percent dissociation = amount dissociated (mol/l) 100 = final [H ] 100 initial concentration (mol/l) initial [HA] For a weak acid, percent dissociation (or ionization) increases as the acid becomes more dilute. Equilibrium shifts to the right. BASES The hydroxides of Group I and IIA metals are all strong bases. The Group IIA hydroxides are not very soluble. This property allows some of them to be used effectively as stomach antacids. Ex. Calculate the [OH ], [H ], and ph of a 0.0100 M solution of NaOH. NaOH is a strong base. [OH ] = 0.0100 M [H ] = 1 10 14 /1 10 2 = 1 10 12 M ph = log 1.00 10 12 = 12.000 Weak bases (bases without OH ) react with water to produce a hydroxide ion. Common examples of weak bases are ammonia (NH 3 ), methylamine (CH 3 NH 2 ), and ethylamine (C 2 H 5 NH 2 ). B(aq) H 2 O(l) BH (aq) OH (aq) base acid conjugate conjugate acid base NH 3 H 2 O NH 4 OH base acid conjugate conjugate acid base The lone pair on N forms a bond with an H. Most weak bases involve N. Acid/Base/Salt Equilibria Notes pg.5 Kristen Jones 6/21/2013

Base dissociation constant (K b ) K b = [BH ][OH ] K b = [NH 4 ][OH ] [B] [NH 3 ] Determination of the ph of a weak base is very similar to the determination of the ph of a weak acid. Follow the same steps. Remember, however, that x is the [OH ] and taking the negative log of x will give you the poh and not the ph! Ex. Calculate the [OH ] and the ph for a 15.0 M NH 3 solution. The K b for NH 3 is 1.8 10 5. R NH 3 H 2 O NH 4 OH I 15.0 --- 0 0 C x --- x x E 15.0x --- x x K b = 1.8 10 5 = x 2 x 2 x = 1.6 10 2 = [OH ] poh = log 1.6 10 2 = 1.78 ph = 141.78 = 12.22 15.0x 15.0 Determination of the ph of Salts Neutral Salts- Salts that are formed from the cation of a strong base and the anion from a strong acid form neutral solutions when dissolved in water. Ex. NaCl, KNO 3 Acid Salts- Salts that are formed from the cation of a weak base and the anion from a strong acid form acidic solutions when dissolved in water. Ex. NH 4 Cl The cation hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution. The anion is simply a spectator ion. NH 4 H 2 O H 3 O NH 3 strong acid weak base Basic Salts- Salts that are formed from the cation of a strong base and the anion from a weak acid form basic solutions when dissolved in water. Ex. NaC 2 H 3 O 2 The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution. The cation is a spectator ion. C 2 H 3 O 2 H 2 O OH HC 2 H 3 O 2 strong base weak acid When determining the exact ph of salt solutions, we can use the K a of the weak acid formed to find the K b of the salt or we can use the K b of the weak base formed to find the K a of the salt. K a K b = K w Acid/Base/Salt Equilibria Notes pg.6 Kristen Jones 6/21/2013

High ph Low ph General Order of ph Strong Bases (SB) Weak Bases (WB) Basic Salts (BS) Neutral Salts (NS) Acidic Salts (AS) Weak Acids (WA) Strong Acids (SA) When asked to rank a mixture of compounds by ph, label each as shown above and then arrange them. Arrange the following 0.1 M solutions in order of decreasing ph. NaOH NaCN KCl CH 3 NH 2 H 3 PO 4 NH 4 NO 3 HBr Ex. Calculate the ph of a 0.15 M solution of sodium acetate. The K a of acetic acid is 1.8 10 5. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (acetic acid) and thus forms a basic solution. The acetate ion hydrolyzes to produce acetic acid and hydroxide ions. R C 2 H 3 O 2 H 2 O HC 2 H 3 O 2 OH I 0.15M --- 0 0 C x x x E 0.15 x x x K b = [HC 2 H 3 O 2 ][OH ] K b = K w = 1 10 14 = 5.6 10 10 [C 2 H 3 O 2 ] K a 1.8 10 5 5.6 10 10 = x 2 x 2 x = 9.2 10 6 0.15x 0.15 [OH ] = 9.2 10 6 poh = log 9.2 10 6 = 5.04 ph = 14.005.04 = 8.96 Acidic and Basic Oxides When metallic (ionic) oxides dissolve in water they produce a metallic hydroxide (basic solution). When nonmetallic (covalent) oxides dissolve in water they produce a weak acid (acidic solution). CaO H 2 O Ca(OH) 2 CO 2 H 2 O H 2 CO 3 Salts of Highly Charged Metals Salts that contain a highly charged metal ion produce an acidic solution. AlCl 3 6H 2 O Al(H 2 O) 6 3 3Cl Al(H 2 O) 6 3 Al(H 2 O) 5 (OH) 2 H The higher the charge on the metal ion, the stronger the acidity of the hydrated ion. The electrons are pulled away from the O H bond and toward the positively charged metal ion. FeCl 3 and Al(NO 3 ) 3 also behave this way. Acid/Base/Salt Equilibria Notes pg.7 Kristen Jones 6/21/2013

REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS Common ion effect- The addition of an ion already present (common) in a system causes equilibrium to shift away from the common ion. For example, the addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because of additional chloride ions. This can be explained by the use of LeChatelier's Principle. NaCl(s) Na Cl The addition of a common ion to a solution of a weak acid makes the solution less acidic. HC 2 H 3 O 2 H C 2 H 3 O 2 If we add NaC 2 H 3 O 2 to the above system, equilibrium shifts to undissociated HC 2 H 3 O 2, raising ph. The new ph can be calculated by putting the concentration of the anion into the K a equation and solving for the new [H ]. Buffered solution-a solution that resists changes in ph when hydroxide ions or protons are added. A buffer solution usually consists of a solution of a weak acid and its salt or a weak base and its salt. Ex. HC 2 H 3 O 2 /C 2 H 3 O 2 buffer system Addition of strong acid: H C 2 H 3 O 2 HC 2 H 3 O 2 (Easiest way to show a strong acid is to use H ) Addition of strong base: OH HC 2 H 3 O 2 H 2 O C 2 H 3 O 2 (Easiest way to show a strong base is to use OH ) NH 3 /NH 4 buffer system Addition of strong acid: H NH 3 NH 4 Addition of strong base: OH NH 4 NH 3 H 2 O Buffer capacity- The amount of acid or base that can be absorbed by a buffer system without a significant change in ph. In order to have a large buffer capacity, a solution should have large concentrations of both buffer components. One way to calculate the ph of a buffer system is with the Henderson-Hasselbach equation. ph = pk a log [base] ph = pk a log [A - ] [acid] [HA] For a particular buffering system, all solutions that have the same ratio of [A - ]/[HA] have the same ph. Optimum buffering occurs when [HA] = [A - ] and the pk a of the weak acid used should be as close as possible to the desired ph of the buffer system. The Henderson-Hasselbach (HH) equation needs to be used cautiously. It is sometimes used as a quick, easy equation to plug numbers into. A K a or K b problem requires a greater understanding of the factors involved and can always be used instead of the HH equation. Hints for Solving Buffer Problems: 1. Determine major species involved initially. 2. If chemical reaction occurs, write equation and solve stoichiometry in moles, then change to molarity. 3. Write equilibrium equation. 4. Set up equilibrium expression (K a or K b ) or HH equation. 5. Solve. 6. Check logic of answer. Acid/Base/Salt Equilibria Notes pg.8 Kristen Jones 6/21/2013

Ex. A solution is 0.120 M in acetic acid and 0.0900 M in sodium acetate. Calculate the [H ] at equilibrium. The K a of acetic acid is 1.8 10 5. R HC 2 H 3 O 2 H C 2 H 3 O 2 I 0.120 0 0.0900 C x x x E 0.120x x 0.0900 x K a = x (0.0900 x) x (0.0900) = 1.8 10 5 0.120x 0.120 x = 2.4 10 5 M [H ] = 2.4 10 5 Using the Henderson-Hasselbach equation: pk a = log 1.8 10 5 = 4.74 ph = 4.74 log (0.0900/0.120) = 4.62 [H ] = 10 (4.62) = 2.4 10 5 M Ex. Calculate the ph of the above buffer system when 100.0 ml of 0.100 M HCl is added to 455 ml of solution. 0.100 L HCl 0.100 M = 0.0100 mol H 0.455 L C 2 H 3 O 2 0.0900 M = 0.0410 mol C 2 H 3 O 2 0.455 L HC 2 H 3 O 2 0.120 M = 0.0546 mol HC 2 H 3 O 2 H C 2 H 3 O 2 HC 2 H 3 O 2 Before 0.0100 mol 0.0410 mol 0.0546 mol Change 0.010 0.0100 0.0100 After 0 0.0310 mol 0.0646 mol Weak acid and its conj. base left Work K a! 0.0310 mol acetate / 0.555 L solution = 0.0559 M acetate 0.0646 mol acetic acid/0.555 L solution = 0.116 M acetic acid R HC 2 H 3 O 2 H C 2 H 3 O 2 I 0.116 M 0 0.0559 M C x x x E 0.116x x 0.0559 x K a = 1.8 10 5 = x(0.0559x) x(0.0559) 0.116 x 0.116 x = 3.74 10 5 M ph = 4.43 Acid/Base/Salt Equilibria Notes pg.9 Kristen Jones 6/21/2013

Acid-Base Titrations titrant-solution of known concentration (in buret) The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumed (stoichiometric point or equivalence point). ph or titration curve -plot of ph as a function of the amount of titrant added. Types of Acid-Base Titrations: 1. Strong acid-strong base Simple reaction H OH H 2 O The ph is easy to calculate because all reactions go to completion. At the equivalence point, the solution is neutral. Ex. 100.0 ml of 1.00 M HCl is titrated with 0.500 M NaOH. Calculate the [H ] after 50.0 ml of base has been added. 0.1000 L 1.00 M = 0.100 mol H 0.0500 L 0.500 M = 0.0250 mol OH H OH H 2 O Before 0.100 mol 0.0250 mol 0 Change 0.0250 0.0250 0.0250 After 0.0750 mol 0 0.0250 mol 0.0750 mol H / (0.100 L 0.0500 L) = 0.500 M H Calculate the [H ] after 200 ml of base has been added. 0.200 L 0.500 M = 0.100 mol OH H OH H 2 O Before 0.100 mol 0.100 mol 0 Change 0.100 0.100 0.100 After 0 0 0.100 mol [H ] is not zero. The [H ] of pure water is 1.0 10 7, therefore ph = 7.00. Calculate the ph after 300 ml of base has been added. 0.300 L 0.500 M = 0.150 mol OH H OH H 2 O Before 0.100 mol 0.150 mol --- Change 0.100 0.100 ---- After 0 0.050 mol --- [OH ]= 0.050 mol/0.400 L = 0.125 M OH poh = 0.913 ph = 13.097 2. Weak acid - strong base The reaction of a strong base with a weak acid is assumed to go to completion. Before the equivalence point, the concentration of weak acid remaining and the conjugate base formed are determined. At halfway to the equivalence point, ph = pk a. At the equivalence point, a basic salt is present and the ph will be greater than 7. After the equivalence point, the strong base will be the dominant species and a simple ph calculation can be made after the stoichiometry is done. Acid/Base/Salt Equilibria Notes pg.10 Kristen Jones 6/21/2013

The ph Curve for the Titration of 50.0 ml of 0.100 M HC2H3O2 with 0.100 M NaOH Ex. 30.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M HF. (K a of HF = 7.2 10 4 ) Determine the ph of the final solution. 0.0300 L 0.10 M = 0.00300 mol OH 0.0500 L 0.10 M = 0.00500 mol HF Stoichiometry OH - HF H 2 O F - Before 0.00300 mol 0.00500 mol --- 0 Change 0.00300 0.00300 --- 0.00300 After 0 0.00200 mol --- 0.00300 mol 0.00200 mol/(0.030l 0.050L) = 0.0250 M HF 0.00300 mol/(0.030l 0.050L) = 0.0375 M F Weak Acid and Conj Base left Work K a! Acid/Base/Salt Equilibria Notes pg.11 Kristen Jones 6/21/2013

Equilibrium R HF H F I 0.025 0 0.0375 C x x x E 0.025 x x 0.0375 x K a = 7.2 10 4 = x (0.0375 x) x(0.0375) 0.0250x 0.0250 x = 4.8 10 4 [H ] = 4.8 10 4 ph = 3.32 Ex. 50.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M HF. (K a of HF = 7.2 10 4 ) Determine the ph of the final solution. 0.050 L 0.10 M = 0.0050 mol OH 0.050 L 0.10 M = 0.0050 mol HF Stoichiometry OH HF H 2 O F Before 0.00500 mol 0.00500 mol 0 0 Change 0.00500 0.00500 ----- 0.00500 After 0 0 ------ 0.00500 mol 0.00500 mol/(0.050 L 0.050 L) = 0.0500 M F - Equilibrium R F H 2 O HF OH I 0.0500 ---- 0 0 C x ----- x x E 0.0500x ---- x x Only weak base left Work K b! K b for F = 1.0 10 14 /K a 1.4 10 11 = [HF][OH ] = x 2 x 2 [F ] 0.0500x 0.0500 x = 8.4 10 7 M [OH - ] =8.4 10 7 poh = 6.08 ph = 7.92 Ex. 60.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M HF. (K a of HF = 7.2 10 4 ) Determine the ph of the final solution. 0.0600 L 0.10 M OH = 0.006 mol OH 0.0500 L 0.10 M HF = 0.0050 mol HF Acid/Base/Salt Equilibria Notes pg.12 Kristen Jones 6/21/2013

Stoichiometry OH HF H 2 O F Before 0.00600 mol 0.00500 mol 0 0 Change 0.00500 0.00500 ----- 0.00500 After 0.00100 mol 0 ----- 0.00500 mol [OH ] = 0.00100 mol/0.110 L = 9.09 10 3 M poh = 2.04 ph = 11.96 Strong base and weak base left Ignore weak base! Weak base - Strong acid Before the equivalence point, a weak base equilibria exists. Calculate the stoichiometry and then the weak base equilibria. At the equivalence point, an acidic salt is present and the ph is below 7. After the equivalence point, the strong acid is the dominant species. Use the [H ] to find the ph. The ph Curve for the Titration of 100.0 ml of 0.050 M NH3 with 0.10 M HCl Acid/Base/Salt Equilibria Notes pg.13 Kristen Jones 6/21/2013

Ex. Calculate the ph when 100.0 ml of 0.050 M NH 3 is titrated with 10.0 ml of 0.10 M HCl. K b of NH 3 = 1.8 10 5 0.100 L 0.050 M = 0.0050 mol NH 3 0.010 L 0.10 M = 0.0010 mol H NH 3 H NH 4 Before 0.0050 mol 0.0010 mol 0 Change 0.0010 0.0010 0.0010 After 0.0040 mol 0 0.0010 mol 0.0010 mol/0.110 L = 9.09 10 3 M NH 4 0.0040 mol/0.110 L = 3.64 10 2 M NH 3 R NH 3 H 2 O NH 4 OH I 0.0364 ----- 0.00909 0 C x x x E 0.0364x 0.00909x x K b = 1.8 10 5 = (0.00909x)x (0.00909)x 0.0364x 0.0364 x = [OH ]= 7.21 10-5 poh = 4.14 ph = 9.86 Ex. Calculate the ph when 100.0 ml of 0.050 M NH 3 is titrated with 50.0 ml of 0.10 M HCl. K b of NH 3 = 1.8 10 5 0.100 L 0.050 M = 0.0050 mol NH 3 0.050 L 0.10 M = 0.0050 mol H NH 3 H NH 4 Before 0.0050 mol 0.0050 mol 0 Change 0.0050 0.0050 0.0050 After 0 0 0.0050 mol 0.0050 mol/ 0.150 L = 0.0333 M NH 4 R NH 4 H 2 O NH 3 H 3 O I 0.0333 ----- 0 0 C x x x E 0.0333x x x K a for NH 4 = (1.0 10 14 /K b for NH 3 ) = 5.56 10 10 5.56 10 10 = [NH 3 ][H 3 O ] x 2 x 2 [NH 4 ] 0.0333x 0.0333 x = 4.30 10 6 ph = 5.37 Acid/Base/Salt Equilibria Notes pg.14 Kristen Jones 6/21/2013

Ex. Calculate the ph when 100.0 ml of 0.050 M NH 3 is titrated with 60.0 ml of 0.10 M HCl. 0.100 L 0.050 M = 0.0050 mol NH 3 0.060 L 0.10 M = 0.0060 mol H NH 3 H NH 4 Before 0.0050 mol 0.0060 mol 0 Change -0.0050-0.0050 0.0050 After 0 0.0010 mol 0.0050 mol Strong acid weak acid Ignore weak acid! 0.0010 mol/0.160 L = 6.25 10-3 M H 3 O ph = 2.20 Titration Curve for Polyprotic Acids Acid-Base Indicators end point- point in titration where indicator changes color When choosing an indicator, we want the indicator end point and the titration equivalence point to be as close as possible. Since strong acid-strong base titrations have a large vertical area, color changes will be sharp and a wide range of indicators can be used. For titrations involving weak acids or weak bases, we must be more careful in our choice of indicator. Indicators are usually weak acids, HIn. They have one color in their acidic (HIn) form and another color in their basic (In ) form. A very common indicator, phenolphthalein, is colorless in its HIn form and pink in its In form. It changes color in the range of ph 8-10. Usually 1/10 of the initial form of the indicator must be changed to the other form before a new color is apparent. The useful range of an indicator is usually its pk a ±1. When choosing an indicator, determine the ph at the equivalence point of the titration and then choose an indicator with a pk a close to that. Acid/Base/Salt Equilibria Notes pg.15 Kristen Jones 6/21/2013