Presentation on theme: "15 - 1 Brønsted-Lowry Definitions The Ion Product for Water The pH and Other “p” Scales Aqueous Solutions of Acids and Bases Hydrolysis The Common Ion."— Presentation transcript:
Brønsted-Lowry Definitions The Ion Product for Water The pH and Other “p” Scales Aqueous Solutions of Acids and Bases Hydrolysis The Common Ion Effect Buffer Solutions Indicators and Titrations Polyprotic Acids Acids and Bases Topic # 14
Arrhenius definitions Acid AcidAnything that produces hydrogen ions in a water solution. HCl (aq) H + + Cl - Base BaseAnything that produces hydroxide ions in a water solution. NaOH (aq) Na + + OH - Arrhenius definitions are limited to aqueous solutions.
Brønsted-Lowry definitions Expands the Arrhenius definitions Acid AcidProton donor Base BaseProton acceptor This definition explains how substances like ammonia can act as bases. NH 3 (g) + H 2 O (l) NH OH -
Brønsted-Lowry definitions Strong acids and bases considered to ionize completely. HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) NaOH (aq) + H 2 O (l) Na + (aq) + OH - (aq) Weak acids and bases do not ionize completely. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) +C 2 H 3 O 2 - (aq) NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)
Brønsted-Lowry definitions When a Brønsted-Lowry acid dissolves in water, the water acts as a base. HC 2 H 3 O 2 (aq) + 2H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Acid Base H 3 O + (aq) is called the hydronium ion. However, the exact number of water molecules required to hydrate a proton is not known.
Brønsted-Lowry definitions Conjugate acid-base pairs. Acids and bases that are related by loss or gain of H + as H 3 O + and H 2 O. Examples. Examples.AcidBase H 3 O + H 2 O HC 2 H 3 O 2 C 2 H 3 O 2 - NH 4 + NH 3 H 2 SO 4 HSO 4 - HSO 4- SO 4 2-
Common acids and bases Acids Acids Formula Molarity* nitricHNO 3 16 hydrochloric HCl 12 sulfuricH 2 SO 4 18 aceticHC 2 H 3 O 2 18Bases ammoniaNH 3 (aq) 15 sodium hydroxideNaOH solid *undiluted.
Common Acids and bases AcidicBasic Acidic Basic Citrus fruitsBaking soda Aspirin Detergents Coca Cola Ammonia cleaners VinegarTums and Rolaids Vitamin CSoap
amphoteric Water is an amphoteric substance that can act either as an acid or a base, HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) acid base acid base H 2 O (l) + NH 3 (aq) NH 4 + (aq) + OH - (aq) acid base acid base
Auto-ionization of water Auto-ionization Auto-ionization When water molecules react with each another to form ions. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) (10 -7 M) (10 -7 M) K w = [ H 3 O + ] [ OH - ] = 1.0 x at 25 o C Note: Note: [H 2 O] is not included in expression because H 2 O is a pure liquid! ion product of water ion product of water
Ion product for water K w is a temperature dependent equilibrium constant. We commonly use 25 o C as the standard temperature. Temperature, o C K w x x x x x Temperature, o C K w x x x x x
Autoionization of water [H + ] and [OH - ] are always present in aqueous solutions. Only for a neutral solution are they at the same concentration. Neutral solution.Neutral solution. [H + ] = M = [OH - ] Acidic solution.Acidic solution. [H + ] > M > [OH - ] Basic solution.Basic solution. [H + ] < M < [OH - ]
pH and other “p” scales We need to measure and use acids and bases over a very large concentration range. And many times knowing those concentrations is very important. pH and pOH are systems to keep track of these very large ranges. pH = -log[H 3 O + ] pOH = -log[OH - ] pH + pOH = 14
pH calculations Determine the following. pH = -log[H + ] pH of 6.7 x M H + = 2.2 pH of 5.2 x M H + = 11.3 [H + ], if the pH is 4.5 = 3.2 x M H +
pOH examples Determine the following. pOH = -log[OH - ] = 14 - pH pOH of 1.7 x M NaOH pOH = 3.8 pH = 10.2 pOH of 5.2 x M H + pH = 11.3 pOH = 2.7 [OH - ], if the pH is 4.5 pOH = 9.5 [OH - ] = 3.2 x M
pH scale A logarithmic scale used to keep track of the large changes in [H + ] M M 1 M Very Neutral Very Basic Acidic When you add an acid, the pH gets smaller. When you add a base, the pH gets larger.
pH of some common materials Substance pH 1 M HCl0.0 Gastric juices Lemon juice Classic Coke2.5 Coffee5.0 Pure Water7.0 Blood Milk of Magnesia 10.5 Household ammonia M NaOH 14.0 Substance pH 1 M HCl0.0 Gastric juices Lemon juice Classic Coke2.5 Coffee5.0 Pure Water7.0 Blood Milk of Magnesia 10.5 Household ammonia M NaOH 14.0
Acid and Base Strength Strong acids Strong acids Ionize completely in water. HCl, HBr, HI, HClO 3, HNO 3, HClO 4, H 2 SO 4. Weak acids Weak acids Partially ionize in water. Most acids are weak. Strong bases Strong bases Ionize completely in water. Strong bases are metal hydroxides - NaOH, KOH Weak bases Weak bases Partially ionize in water. no Ka ! 100% Ionization! no Kb ! 100% Ionization!
LiOHlithium hydroxide NaOHsodium hydroxide KOHpotassium hydroxide RbOHrubidium hydroxide CsOHcesium hydroxide *Ca(OH) 2 calcium hydroxide *Sr(OH) 2 strontium hydroxide *Ba(OH) 2 barium hydroxide These bases are considered to be strong. * Completely dissociated in solutions of 0.01 M or less. These are insoluble bases which ionize 100%. The other five in the list can easily make solutions of 1.0 M and are 100% dissociated at that concentration.
Acid and Base Strength For strong acids and bases, we can directly calculate the pH or pOH if we know the molar concentration.Examples M HCl would produce 0.15 M H +. pH = -log(0.15) = 0.82 pOH = pH = = M NaOH produces M OH -. pOH = -log(0.052) = 1.28 pH = pOH = = 12.72
Acid dissociation constant, K a The ionization of a weak acid can be expressed as an equilibrium. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) The strength of a weak acid is related to its equilibrium constant, K a. K a = We omit water. It’s a pure liquid. We omit water. It’s a pure liquid. [A - ][H 3 O + ] [HA]
Weak acid equilibria Example Determine the pH of a 0.10 M benzoic acid solution at 25 o C if K a = 6.28 x HBz (aq) + H 2 O (l) H 3 O + (aq) + Bz - (aq) The first step is to write the equilibrium expression. K a = [H 3 O + ][Bz - ] [HBz]
Weak acid equilibria HBzH 3 O + Bz - Initial conc., M Change, M -x x x Eq. Conc., M x x x [H 3 O + ] = [Bz - ] = x We’ll assume that [Bz - ] is negligible compared to [HBz]. The contribution of H 3 O + from water is also negligible.
Weak acid equilibria Solve the equilibrium equation in terms of x K a = 6.28 x = x = (6.28 x )(0.10) = M pH= 2.60 x
Dissociation of bases, K b The ionization of a weak base can also be expressed as an equilibrium. B (aq) + H 2 O (l) BH + (aq) +OH - (aq) The strength of a weak base is related to its equilibrium constant, K b. K b = [OH - ][BH + ] [B]
K a and K b values For weak acids and bases K a and K b always have values that are smaller than one. Acids with a larger K a are stronger than ones with a smaller K a. Bases with a larger K b are stronger than ones with a smaller K b. For a conjugate acid:base pair, pK a + pK b = 14 = pK w Most acids and bases are weak.
Ionization constants at 25 o C. AcidpK a pK b Acetic acid Ammonium ion Benzoic acid Formic acid Lactic acid Phenol AcidpK a pK b Acetic acid Ammonium ion Benzoic acid Formic acid Lactic acid Phenol
Hydrolysis Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction. CH 3 COO - (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) This type of reaction is given a special name.Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH -. The reaction of a cation with water to produce the conjugate base and H 3 O +.
Hydrolysis Al(H 2 O) 6 3+ (aq) + H 2 O (l) Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) These metal ions are able to pull electrons from the H-OH bond. If the pull is strong enough, water is split. One of the surrounding water molecules will take on the H + and form H 3 O +. There is only a small # of cations with a high enough positive charge that can act as acids in hydrolysis.
Common ion effect (Just an example of Le Châtelier’s principle.) Common ion effect The shift in equilibrium caused by the addition of an ion formed from the solute. Common ion An ion that is produced by more than one solute in an equilibrium system. Adding the salt of a weak acid to a solution of weak acid is an example of this.
Common ion effect Example What is the concentration of hydrogen ions in a solution formed by adding mol of sodium acetate to one liter of M acetic acid (HAc). (Ac = C 2 H 3 O 2 - ) (assume that the volume of the solution does not change when the salt is added) [Ac - ] [H + ] [HAc] K a = = 1.7 x 10 -5
Common ion effect HAcAc - H + Init. Conc., M Change, M -x+x+x Eq. Conc., M x x x To solve for x, substitute the equilibrium values into the expression. K a [HAc] [Ac - ] (1.7 x ) (0.099-x) ( x) x = =
Common ion effect If we assume that x is negligible compared to and 0.097, we can simplify the problem to: x = x = 1.7 x = [H + ] Note. Note. When a solution contains equal concentrations of an acid and its conjugate base, [H + ] = K a. (1.7 x ) (0.099) (0.097)
Salt Hydrolysis A salt is formed between the reaction of an acid and a base. HCl + NaOH NaCl + H 2 O Usually, a neutral salt is formed when a strong acid and a strong base are neutralized in the reaction. Ions of Neutral Salts CATIONS Na+K+Rd+Cs+ Mg +2 Ca +2 Sr +2 Ba +2 ANIONS Cl-Br-I- ClO 4 -BrO 4 -ClO 3 -NO 3 -
In this NEUTRAL environment, all the ions are spectator ions from the strong acid and strong base reaction. There are no insoluble particles. These ions have little tendency to react with water, they just stay dissociated into 100% ions. So, salts consisting of these ions are called neutral salts and the resulting solution has a pH of 7. For example: NaCl, KNO 3, CaBr 2, CsClO 4 are neutral salts.
Acidic Ions NH 4 +Al +3 Pb +2 Sn +2 TransitionMetalIons HSO 4 - H 2 PO 4 - Basic Ions F-C2H3O2-C2H3O2-NO 2 -HCO 3 - CN - CO 3 -2 S -2 SO 4 -2 HPO 4 -2 PO 4 -3 When weak acids and weak bases react, the strength of the stronger ion conjugate acid or conjugate base in the salt determines the pH of its solutions.
The resulting salt solution can be acidic, neutral or basic. A salt formed between a strong acid and a weak base is an acid salt, for example NH 4 Cl: ( HCl + NH 3 NH 4 Cl ) Strong weak The NH 4 Cl is soluble and reacts with H 2 O: NH H 2 O NH 3 + H 3 O+ The act of the salt particles reacting with water molecules in the ‘salt solution’ is HYDROLYSIS.
A salt formed between a weak acid and a strong base is a basic salt, for example NaCH 3 COO. (CH 3 COOH + NaOH NaCH 3 COO + H 2 O) CH 3 COO - + H 2 O CH 3 COOH + OH - pH > 7
Deciding whether a salt is acidic, basic or neutral! ******Decide which ion in the salt could produce a stronger acid or stronger base. Example: NaF ---- NaOH or HF NaOH is the stronger base so NaF is a basic salt NaClO NaOH or HClO 4 BOTH are strong so neutral salt ! Fe(NO 3 ) Fe(OH) 2 or HNO 3 HNO 3 is the stronger compound Fe(NO 3 ) 2 is an acidic salt
What if both ions are from weak acids and weak bases? Salt: NH 4 NO 2 NH 4 OH and HNO 2 weak base and weak acid Can compare K b for NH 4 OH and K a for HNO 2 : K b NH 4 OH = 1.6 x10 -5 K a for HNO 2 = 4.5 x K a > K b so the salt is acidic!
Another example: ( qt.1h. on Salt Hydrolysis WS) (NH 4 ) 2 CO NH 4 OH and HCO 3 - weak base and weak acid ion K b for NH 4 OH = 1.6 x K a2 for H 2 CO 3 = 4.8 x So (NH 4 ) 2 CO 3 is basic!
Calculate the pH of a M solution of KCN. ( K a for HCN is 5.8 x ) Calculations !!! CN - + H 2 O HCN + OH - 0.5M +x +x K b = [HCN] [OH-] [CN-] K b = KwKaKwKa 1 x x = = 1.7 x10 -5 Salt is a basic salt, so need to use K b …. - ion makes OH- !
x10 -5 = x –x x = 2.9 x M pOH = -log(2.9 x ) = = = pH /
Buffers Solutions that resist pH change when small amounts of acid or base are added. Two types weak acid and its salt weak base and its salt HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Add OH - Add H 3 O + shift to right shift to left Based on the common ion effect. Attacks HA
A solution of a weak acid and its’ salt! Acetic Acid and Sodium Acetate: HC 2 H 3 O 2 + NaC 2 H 3 O 2 Mix these two compounds in water gives a solution with this organization: HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Concentration of the C 2 H 3 O 2 - would be greater than just a solution of low soluble HC 2 H 3 O 2 !!!! Buffers
Ethanoic acid and ethanoate: CH 3 CO H 3 O + HCH 3 CO 2 + H 2 O The ethanoate ions will be able to absorb any excess acid that is added HCH 3 CO 2 + OH - CH 3 CO H 2 O The ethanoic acid will be able to absorb any excess base that is added Both: CH 3 CO 2 - AND CH 3 CO 2 H are present in solution How does a buffer work?
Buffers The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two. Calculating pH of an acidic buffer use: pH = pK a + log Calculating the pH of a basic buffer use: pH = 14 - ( pK b + log ) [HA] [A - ] [HA] Henderson-Hasselbalch equation
Buffers The beauty of a buffer…. A total of 100 ml of a 1.0 M HCl is add in 10 ml increments to 100 ml of each of the following: Solution A: Pure water - pH = 7 (no acid in it ! ) Solution B: A solution containing 1.0 M HA (acid in it ), 1.0 M A - (conjugate base in it ) with a pK a of 7.0 Calculate the pH after each addition.
Buffers Initially, each sample is at pH 7.00 After adding 10 ml of 1.0 M HCl we have: Pure water [H 3 O + ] = = pH = 1.04 From pH 7 to pH 1.04….. This is a pretty big jump! (10 ml)(1.0 M) (110 ml)
Buffers Add 10 ml of 1.0 M HCl to our buffered system..10 moles +.01 moles.10 moles -.01 moles moles moles We started with 0.10 moles of both the acid and conjugate base forms. The HCl in our first 10 ml can be expected to react with the conjugate base, converting it to the acid. After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form. CH 3 CO H 3 O + HCH 3 CO 2 + H 2 O
Buffers Checking the pH: Since all we need to be concerned about is the ratio of A - to HA in the 10-90% region then: pH= pK A + log = log = 6.91 A change of only 0.09 (Buffer solution) compared to 5.96 ( in pure water) ! [A - ] [HA] 0.09 mol 0.11 mol
Buffers ml HCl pH added unbuffered buffered Matter of fact……..
Buffers ml HCl added pH buffered unbuffered
Indicators Acid-base indicators are highly colored weak acids or bases. HIndic Indic - color 1color 2 color 1 color 2 They may have more than one color transition. Example. Example. Thymol blue Red - Yellow - Blue One of the forms may be colorless - phenolphthalein (colorless to pink)
Indicators Indicator color change. Indicators are commonly used to detect the endpoint of an acid-base titration. The indicator should not start to undergo a color change until about one pH unit after the equivalence point.
Indicators Selection of an indicator. Under these conditions, the transition range for our indicator is pK a + 1. Ideally then, you want your indicator pK a to be one above the pK a for an acid sample. For bases, it should be one lower. You can seldom find a ‘perfect’ indicator so should use one that is a maximum of one unit away.
Indicators pH transition Indicator range color Bromophenol Blue yellow-blue Methyl Orange red-yellow Methyl Red red-yellow Bromothymol Blue yellow-blue Cresol Purple yellow-purple Phenolphthalein colorless - pink Thymolphthaleine colorless - blue Alizerin Yellow GG yellow - red
Indicator examples Acid-base indicators are weak acids that undergo a color change at a known pH. phenolphthalein pH
Indicator examples methyl red bromthymol blue
Titrations revisited Methods based on measurement of volume. If the concentration of an acid is known, the amount of a base can be found. If we know the concentration of the base, then we can determine the amount of acid. All that is needed is some calibrated glassware and either an indicator or pH meter.
Titrations Buret Buret - volumetric glassware used for titrations. It allows you to add a known amount of your titrant to the solution you are testing. If a pH meter is used, the equivalence point can be measured. An indicator will give you the endpoint.
Titrations Note the color change which indicates that the ‘endpoint’ has been reached. StartEnd
Titration curves Acid-base titration curve A plot of the pH against the amount of acid or base added during a titration. Plots of this type are useful for visualizing a titration. It also can be used to show where an indicator undergoes its color change.
Titration curves pH Equivalence Point % titration or ml titrant Buffer region Overtitration Indicator Transition Four regions of titration curve
Titration curves Strong acid titrated with a strong base. This is pretty straight forward since the net reaction is: H 3 O + + OH - 2H 2 O Prior to equivalence point The pH indicates the amount of sample present after accounting for dilution. [H 3 O + ] = moles acid - moles base total volume in liters
Titration curves Strong acid titrated with a strong base. Equivalence point [H 3 O + ] = [OH - ] pK W = 14 = pH + pOH pH = 7.00 So the equivalence point for strong acid/base problems is always at pH=7.00
Titration curves Strong acid titrated with a strong base. Overtitration Past the equivalence point, we don’t have any acid remaining. All that we are doing is diluting our titrant by adding more solution volume and more OH- ions. [OH - ] = pH = 14 - pOH moles excess total volume in liters
Titration curves Example. Construct a titration curve for the titration of 100 ml of 0.10 M HCl with 0.10 M NaOH 0% titration A. 0% titration pH = -log[0.10] = 1 B. After 10 ml NaOH [H 3 O + ] = = M, pH = 1.09 (100 ml)(0.10 M) - (10 ml)(0.10 M) 100 ml + 10 ml
Titration curves ml titrant total ml[H 3 O + ]pH
Titration curves ml NaOH pH Let’s add 10ml more!
Titration curves ml NaOH pH
Titration curves Reached the Equivalence point or 100 % titration or 100 % titration (100ml acid)(0.10M acid) = (100ml base)( 0.10M base) pH MUST be 7 1 x = (1 x10 -7 ) 2 Note that for the first 90 ml of our titration, we only saw a change of 1.28 pH units. Now we have a jump of 4.72 pH units.
Titration curves Anything over the equilvalence point is Overtitration: (no more acid there!) Describing the dilution of our titrant: 10 ml overtitration [OH - ] = 0.10 M = M pOH = 2.32 pH= = (.10M)10 ml 210 ml
Titration curves ml titrant total volume [OH - ] pH Continue the calculations:
Titration curves ml NaOH pH Over titration Just adding more OH-
Titration curves Titration of a strong base with a strong acid. This is not significantly different from our earlier example. If you plot pOH rather that pH, the results would look identical. Typically we still plot pH versus ml titrant so the curves are inverted.
Titration curves pH ml titrant basic sample (base in the beaker) acidic sample (acid in the beaker)
Titration of weak acids or bases Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example. We must be concerned with conjugate acid/base pairs and their equilibria.Example HA + H 2 O H 3 O + + A - acid base
Titration of weak acids or bases First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid. We still have the same four general regions for our titration curve. The calculation will require that you use the appropriate K a or K b relationship. We’ll start by reviewing the type of calculations involved and then work through an example.
Titration of weak acids or bases 0% titration If your sample is an acid then use K a = At this point [H 3 O + ] = [A - ]. You can solve for [H 3 O + ] by using either the quadratic or approximation approach. Finally, calculate the pH. [H 3 O + ][A - ] [HA]
Titration of weak acids or bases 0% titration If your sample is an base then use K b = At this point [OH - ] = [HA]. You can solve for [OH - ] by using either the quadratic or approximation approach. Then determine pH as 14 - pOH. [OH - ][HA] [A - ]
Titration of weak acids or bases % titration In this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our answer. Henderson-Hasselbalch equation A common format for the equilibrium expression used in this region is the Henderson-Hasselbalch equation. We can present it in two forms depending on the type of material we started with.
Titration of weak acids or bases % titration Starting with an acid pH = pK a + log Starting with a base pH = 14 - ( pK b + log ) ( We’re just determining the pOH and then converting it to pH. ) [HA] [A - ] [HA]
Titration of weak acids or bases % titration Another approach that can be taken is to simply use the % titration values. For an acidic sample, you would use: pH = pK a + log % titration % titration
Titration of weak acids or bases % titration These equations have their limits and may break down if: K a or K b > You are working with very dilute solutions. In those cases, you need to consider the equilibrium for water
Titration of weak acids or bases 100% titration - the equivalence point. At the point, we have converted all of our sample to the conjugate form. If your sample was an acid, now solve for the pH using the K b relationship -- do the opposite if you started with a base. Remember that pK a + pK b = 14 You must also account for dilution of your sample as a result of adding titrant.
Titration of weak acids or bases Overtitration (> 100%) These calculations are identical to those covered in our strong acid/strong base example. You simply need to account for the amount of excess titrant added and how much it has been diluted.
Titration of weak acids or bases Example A 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve. For benzoic acid K a = 6.28 x pK a = 4.20
Titration of weak acids or bases Before adding any base: K a = [H 3 O + ] = [A - ] [HA] + [A - ] = 0.10 M (We’ve assumed that [A - ] is negligible compared to [HA].) [H 3 O + ][A - ] [HA]
Titration of weak acids or bases Initial pH, before adding any base: K a = 6.28 x = x = (6.28 x )(0.10) = M pH= 2.60 x
Titration of weak acids or bases adding 10ml increments of base: adding 10ml increments of base: Here we can use the Henderson- Hasselbalch equation in ml titration format pH = pK a + log pH = log (10 / 90) = 3.25 We can calculate other points by repeating this process. (increments of 10ml) (100ml – total added) (Showing the first increment of 10ml)
Titration of weak acids or bases ml added pH Note: At 50 ml titration, pH = pK a Also, the was only a change of 1.91 pH units as we went from 10 to 90 ml titration. (5.15 – 3.24) Note: At 50 ml titration, pH = pK a Also, the was only a change of 1.91 pH units as we went from 10 to 90 ml titration. (5.15 – 3.24)
Titration of weak acids or bases % titration pH
Titration of weak acids or bases 100% titration (EQUILVALENCE POINT) At this point, virtually all of our acid has been converted to the conjugate base - benzoate. Need to use the K b relationship to solve for this point! (have no acid left!) K b = K b = K w / K a = 1.59 x [OH - ] [HA] [A - ]
Titration of weak acids or bases EQUILVALENCE POINT At the equivalence point: [HA] = [OH - ] (We’ve diluted the [A - ] = 0.05 Msample and the total volume at this point is 200 ml) Finally, assume that [benzoic acid] is negligible compared to [benzoate]..10m.200 liters
Titration of weak acids or bases 100% titration (EQUILVALENCE POINT) K b = 1.59 x = x = (1.59 x )(0.050) = 2.81 x M pOH = 5.55pH = 14 - pOH = 8.45 x
Titration of weak acids or bases % titration pH
Titration of weak acids or bases Overtitration All we need to do here is to account for the dilution of our titrant. 10 % overtitration (10 ml excess) [OH - ] = 0.10 M = M pOH = 2.32 pH= = 11.68
Titration of weak acids or bases ml titrant total volume [OH - ] pH This is identical to what we obtained for our strong acid/strong base example
Titration of weak acids % titration pH
Polyprotic acids A number of acids exist that have more that one ionizable hydrogen.Examples H 3 PO 4 Phosphoric Acid H 2 SO 4 Sulfuric Acid H 2 C 2 O 4 Oxalic acid H 2 CO 3 Carbonic acid Each H + will ionize with its own constant. Also, it becomes more difficult to remove subsequent H +.
Stepwise equilibrium Example Example - H 3 PO 4 Each hydrogen is capable of dissociation but not to the same extent at any given pH. H 3 PO 4 H 2 PO 4 - HPO 4 2- PO 4 3- Since the removal of any one H + affects how easily subsequent H + are removed, each step has its own K a value. Ka3Ka3 Ka2Ka2 Ka1Ka1
Stepwise equilibrium [H 3 O + ] [PO 4 3- ] [HPO 4 2- ] K A 3 = [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] K A 2 = [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] K A 1 = Note: [H 3 O + ] is the same for each expression. The relative amounts of each species can be found if the pH is known. The actual amounts can be found if pH and total H 3 PO 4 is known. Note: [H 3 O + ] is the same for each expression. The relative amounts of each species can be found if the pH is known. The actual amounts can be found if pH and total H 3 PO 4 is known.