Lecture 8 Professor Hicks Inorganic Chemistry (CHE152) Making a Buffer Buffers buffers = solutions that resist ph changes act by neutralizing added acid or base made by preparing a solution of a weak acid/base and the salt of its conjugate acid/base most of the weak acid does not dissociate into H 3 O + but it can react with OH most of the weak base does not react to form OHbut it can react with any H 3 O + that is added HC 2 H 3 O 2 HC 2 H 3 O 2 C H H + 2 H 3 O 2 + H + HC 2 H 3 O C 2 2 H 3 O 2 HC 2 H 3 O 2 H + H + HC 2 H 3 O 2 C 2 H 3 O 2 Free H + determines the ph but all the H can react with added OH the overall effect is that the ph will not change as much as water would if a strong acid or base was added
how buffers work against strong acid new HA H 2 O HA + H 3 O+ HA A Added H 3 O + new A how buffers work against strong base H 2 O HA A HA A + H 3 O+ Added HO ICE tables for buffers What is the ph of a buffer that is 0.55 M HC 2 H 3 O 2 and 0.66 M NaC 2 H 3 O 2? HC 2 H 3 O 2 (aq) + H 2 O C 2 H 3 O 2 (aq) + H 3 O + (aq) initial change equil 0.55 0.66 0 x +x +x 0.55x 0.66+x +x 1.76 x 10 5 = 1.76 x 10 5 = (0.66 + x)x (0.55x) 0.66x 0.55 acetic acid K a = 1.76 x 10 5 x is small compared to the molarity of acid and base because they are both weak x = 1.46 x 10 5 = [H 3 O + ] ph = 4.83
ICE tables for buffers Calculate the change in ph when 1.0 ml of concentrated HCl (12.0 M) is added to 99.0 ml of a buffer that is 0.55 M HC 2 H 3 O 2 and 0.66 M NaC 2 H 3 O 2 (the buffer of the previous slide). Calculate the initial molarity of H 3 O + from the added HCl using dilution formula M dil V dil =M conc V conc M dill = M conc V conc V di M dill = 12 M x 0.001 lit 0.10 = 0.12 M Increasing the total volume from 99 to 100 ml changes the molarities of HC 2 H 3 O 2 and HC 2 H 3 O 2 about 1% so we can use the values of 0.55 and 0.66 M. initial change equil 1.76 x 10 5 = HC 2 H 3 O 2 (aq) + H 2 O C 2 H 3 O 2 (aq) + H 3 O + (aq) 0.55 0.66 0.12 x +x +x 0.55x 0.66+x 0.12+x (0.66 + x)(0.12+x) (0.55x) you can t neglect x everywhere otherwise there is no x! you could solve using quadratic formula, or HC 2 H 3 O 2 (aq) + H 2 O C 2 H 3 O 2 (aq) + H 3 O + (aq) initial change equil 0.55 table from 0.66 0.12 x previous slide +x +x 0.55x 0.66+x 0.12+x If all the H 3 O + reacted with C 2 H 3 O 2 HC 2 H 3 O 2 (aq) + H 2 O C 2 H 3 O 2 (aq) + H 3 O + (aq) initial 0.67 0.54 0 change x +x +x equil 0.67x 0.54+x +x 1.76 x 10 5 = (0.54 + x)x (0.67x) 1.76 x 10 5 = 0.54 x 0.67 x= 1.76 105 0.67 0.54 = 2.18 10 5 ph = log( 2.18 10 5 ) = 4.66 the change in ph is 4.66 4.83 = 0.17 Calculate the change in ph when 1.0 ml of concentrated HCl (12.0 M) is added to 99 ml of water. M dil V dil =M conc V conc M dill = M conc V conc V di M dill = 12 M x 0.001 lit 0.10 = 0.12 M Since HCl is a strong acid and this solution only has HCl and water the ph is ph = log( 0.12) = 0.92 the change in ph is 0.92 7.0 = 6.08 compared to 0.17 for the buffered solution!
HendersonHasselbalch Equation used to calculate ph of buffer solutions equation derived from the K a expression input = initial concentrations of the weak acid and conjugate base moles of acid and base can also be used assumes the x is small approximation ph pk a [conjugate base anion] log [weak acid] pk a = log(k a ) initial initial Deriving the HendersonHasselbach Equation [A ][H3O ] Ka HA [HA] [H3O ] Ka [A ] [HA] ph pk a log [A ] ph p K a [A ] log [HA] take log both sides and substitute ph log[h3o pk a log K a use this property of logarithms [HA] [A ] log log [A ] [HA] A is sometime just written as B for base and HA as A for acid ] Example: What is the ph of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2? HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 + H 3 O + [A ] ph pk a log [HA] 0.150 ph 4.187 log 0.050 ph 4.66 ph [H3O ] 10 4.66 5 [H3O ] 10 2.210 2.2 10 0.050 5 100% 0.044% 5% log K a for HC 7 H 5 O 2 = 6.5 x 10 5 pk a log K a 6.5 10 5 4. 187 recognize chemical formula has 2 oxygen atoms it is a carboxylic acid weak acid so using the HHA equation is a valid approximation
Buffering Effectiveness the buffering capacity is the amount of acid or base a buffer can neutralize the buffering range is the ph range the buffer can be effective the effectiveness of a buffer depends on two factors 1) the ratio of base to acid 2) concentrations of acid and base Buffering Capacity [base]/[acid] ratio = 1, greatest capacity against acid and base buffers that need to work mainly against added acid have larger capacity if [base] > [acid] buffers that need to work mainly against added base have largest capacity if [acid] > [base] buffer capacity is always larger when both acid and base concentrations are higher Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer
Buffering Range buffer will be effective when 0.1 < [base]/[acid] < 10 substituting into the HendersonHasselbalch we can calculate the maximum and minimum ph at which the buffer will be effective [A ] ph pk a log [HA] Lowest ph Highest ph ph pka log0.10 ph pka log10 ph pk 1 ph pk 1 a the reasonably effective ph range of a buffer is pk a ±1 when choosing an acid to make a buffer, choose one whose is pk a is within 1 unit of the desired ph a 16 Example: Calculate the ph of a solution prepared so that it was 0.085 M in nitrous acid (HNO 2 ) and 0.10 M in sodium nitrite (NaNO 2 ). This solution is a buffer b/c we have a weak acid and its conjugate bases use the HH eqn 1) look up K a 2) calculate pk a 3) calculate [B]/[A] ph = pk a + log([b]/[a]) B = NO 2 A =HNO 2 K a = 4.6 x 10 4 ph = 3.337 + log (1.17) ph = 3.405 [B]/[A] = 0.10/0.085 = 1.17 pk a = log (4.6 x10 4 ) = 3.337 Example: Calculate the ph of a solution prepared so that it was 0.85 M in nitrous acid (HNO 2 ) and 1.0 M in sodium nitrite (NaNO 2 ). This solution is a buffer bc we have a weak acid and its conjugate bases use the HH eqn 1) look up K a 2) calculate pk a 3) calculate [B]/[A] ph = pk a + log([b]/[a]) B = NO 2 A =HNO 2 K a = 4.6 x 10 4 ph = 3.337 + log (1.17) ph = 3.405 [B]/[A] = 1.0/0.85 = 1.17 pk a = log (4.6 x10 4 ) = 3.337 [H 3 O + ] = 10 3.405 = 3.93 x10 3 % diss = (3.93 x10 3 /0.86) x 100% = 0.04% approximation is OK this buffer has same B/A ratio and therefore same ph as previous example
Selecting a conjugate acid/base pair for a buffer buffer most effective when ph = pk a select an acid with a pk a close to desired ph Example: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with ph 4.25? Chlorous Acid, HClO 2 pk a = 1.95 Nitrous Acid, HNO 2 pk a = 3.34 Formic Acid, HCHO 2 pk a = 3.74 Hypochlorous Acid, HClO pk a = 7.54 pk a of 3.74 is closest to the desired ph of 4.25 Buffer ratio (B/A) a) Which acid and its conjugate base would be the best choice to make a ph 4.0 buffer? b) Calculate the [B]/[A] ratio be for this buffer ph = pk a + log ([B]/[A]) 4.0 = 4.01 + log ([B]/[A]).01 = log ([B]/[A]) 10 0.01 = ([B]/[A]) [B]/[A] = 0.97 by adding a little less base than acid the ph will be lower than the pk a
Titrations burette has solution of known molarity = titrant flask has unknown solution = analyte titrant added until the endpoint is reached at endpoint the titrant has been added in slight excess excess added reacts with an indicator that changes color calculations use the approximation endpoint = equivalence point equivalence point is the point where the number of equivalents of titrant and analyte are equal titrant analyte Indicators indicators change color depending on the ph of the solution indicators are weak acids/bases that get titrated along with the analyte HInd (aq) + H 2 O (l) Ind (aq) + H 3O + (aq) different color added in such small amounts they change the ph negligibly
AcidBase Indicators 25 strong acid with strong base both dissociate completely H 3 O + /OH H 3 O + /OH react completely to form water ph determined by how much is in H 3 O + /OH excess titration of acids 14 weak acid with strong base 1) initial ph is that of weak acid solution 2) in middle of titration solution is a buffer Use HHE 3) at the endpoint acid is converted to its conjugate base in new total volume do eq calc with initial conc of base ph is will be basic 12 ph 10 8 6 4 2 1 2 3 0 0 5 10 15 20 25 30 35 40 45 50 Volume NaOH Added, ml titration of weak acid with strong base 1) initial ph is calculated using standard approximation for a weak acid 2) as the titration proceeds acid is converted into conjugate base 3) towards the middle of titration [B]/[A] 0.10 to 10 the solution is a buffer use HHE 4) at the equivalence point make approximation acid was completely neutralized forming a solution of conjugate base 5) calculate the ph using the standard approximation for hydrolysis reaction of the conjugate base of the acid 6) for steps 4 and 5 you must take account for the dilution due to adding the titrant
If a 10 ml of 0.10 M acetic acid solution is titrated with 0.10 M NaOH calculate: a) The ph before the titration begins. b) the ph 1/3 of the way to the c) the ph 1/2 of the way to the d) the ph at the If a 10 ml of 0.10 M hydrochloric solution is titrated with 0.10 M NaOH calculate: a) The ph before the titration begins. b) the ph 1/3 of the way to the c) the ph 1/2 of the way to the d) the ph at the If a 10 ml of 0.10 M ammonia (NH 3 ) solution is titrated with 0.10 M HCl calculate: a) The ph before the titration begins. b) the ph 1/3 of the way to the c) the ph 1/2 of the way to the d) the ph at the