Lecture 10 Professor Hicks Inorganic Chemistry II (CHE152) ph Scale of [H 3 O + ] (or you could say [H + ]) concentration More convenient than scientific notation ph = log [H 3 O + ] still not sure? take the log of 10 it should be 1 [H 3 O + ] 1
poh Same idea as ph Scale of OH concentration poh = log [OH ] [H 3 O + ] K eq pk eq K eq values vary over a VERY wide range K eq 10 100 and K eq 10 100 As an alternative to using scientific notation often expressed a pk eq values For example K eq for acids K a pk a K eq for bases K b pk b K eq for autoionization of water K w pk w 2
ph, poh, and pk w ph + poh = 14 Why? [H 3 O + ][OH ] = 10 14 1) take log both sides 2) log(a*b) = log(a) + log(b) ph is what most people think in terms of some problems we get a result [OH] or poh use this equation to express it as a ph ph + poh = log(10 14 ) = 14 ph = 14 poh ph fun facts! Bacteria that are not harmful tend to grow in acidic conditions (acidophilus strains) More harmful bacteria tend to grow in basic conditions Blood ph about 7.4 Stomach ph 1.5! Acid rain lowers ph of environment dissolving Al(OH) 3 solid 3
Why is ph (in other words [H + ]) important? H + and OH form from water Life on earth exists in water H + and OH promote numerous reactions as catalysts Their concentration must be carefully controlled a ph change of 0.50 causes irreversible damage change in [ H + ] of only 9.6x10 7!!! ph poh [OH ] [H + ] ph poh = 14 ph ph = 14 poh poh [H + ] = 10 ph ph = log [H + ] [OH ] = 10 poh poh = log [OH ] [H + ] [OH ] = 10 14 [H + ] [H + ] = 10 14 [OH ] [OH ] 4
5
6
7
8
bases NaOH Sr(OH) 2 9
Monoprotic Acids Monoprotic acids acids that provide only one H + (H 3 O + ) Common examples of monoprotic acids are here H 2 SO 4 is a diprotic acid H 3 PO 4 is a triprotic acid We will only cover calculations for monoprotic acids 10
Dissociation of weak acids Calculate [H 3 O + ] at equilibrium for a 0.55 M solution of HF in water. Write a hydrolysis reaction Acid + H 2 O H 3 O + + conj base initial (M) change (M) equilibrium (M) HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) 0.55 0 0 x +x +x 0.55x +x +x K a = [H 3 O + ][F] [HF] to solve for x you could use the quadratic equation a quicker alternative is to look for an approximation look up K a for HF x 2 7.1 x 10 4 = x =??? 0.55x if x is very small compared to 0.55 7.1 x 10 4 = x 2 0.55 x = sqrt{ 0.55* 7.1 x 10 4 } = 0.020 M x = change in [HF] to reach equilibrium, and is also final equilibrium molarities for [H 3 O + ], [F ] Percent Dissociated (aka percent ionized) % Dissociated = x Initial molarity x 100% Initial molarity HF before dissociation dissolve Initial molarity x HF (aq) H + (aq) F (aq) x at equilibrium 11
initial (M) change (M) equilibrium (M) Dissociation of weak acids HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) 0.55 0 0 x +x +x 0.55x +x +x K a = [H 3 O + ][F] [HF] to solve for x you could use the quadratic equation a quicker alternative is to look for an approximation x 2 7.1 x 10 4 = x =??? 0.55x % dissociated is 0.020 is small compared to 0.55 M? 0.020 0.55 if x is very small compared to 0.55 x 100% = 3.6% 7.1 x 10 4 = x 2 0.55 x = sqrt{ 0.55* 7.1 x 10 4 } = 0.020 M 5% error is widely accepted in science Approximation is valid b/c % dissociated is less than 5% Le Chateliers Principle and % dissociated HF + H 2 O F + H 3 O + K a = 7.1 x 10 4 ~100% 90% 10% 40% 50% 60% 70% 80% 1% 30% at infinite dilution percent dissociated disturbance = increase H 2 O (dilute acid) 1.0 M 0.10 M 0.001 M response = decrease [H 2 O] 0.00010 M etc 12
1) Break into groups of 23 each group will be assigned an acid 2) Use the approximation method to estimate the ph and % dissociated. If more than 5% is dissociated at equilibrium then the approximation is not good. Acid Acid Concentration K a [H 3 O + ] diss 1) HF 4.50E01 7.10E04 1.79E02 2) HNO 2 4.23E01 4.50E04 1.38E02 3) C 9 H 8 O 4 2.19E01 3.00E04 8.10E03 4) HCOOH 1.08E01 1.70E04 4.28E03 5) C 6 H 8 O 6 6.83E02 8.00E05 2.34E03 6) C 6 H 5 COOH 6.09E02 6.50E05 1.99E03 7) CH 3 COOH 1.26E02 1.80E05 4.76E04 8) HCN 3.48E07 4.90E10 1.30E08 9) C 6 H 5 OH 1.26E07 1.30E10 4.04E09 13
A 0.123 M solution of a weak acid is found to have a ph of 2.34. What is the K a of this weak acid? What is this acid s pk a value? 14
Calculate the ph of a 0.345 M solution of acetic acid. Confirm that your calculation is valid by calculating the percent dissociated. Ionization of weak bases Calculate ph at equilibrium for a 0.10 M solution of Na 2 CO 3 in water. write hydrolysis reaction base + H 2 O conj acid + OH initial (M) change (M) equilibrium (M) K b = [OH ][HCO 3 ] [CO 3 2 ] look up K b for CO 3 2 1.8 x 10 4 = x 2 0.10x CO 2 3 (aq) + H 2 O (l) HCO 3 (aq) + OH (aq) 0.10 0 0 x +x +x 0.10 x x x 1.8 x 10 4 = x 2 0.10 x = sqrt{ 0.10* 1.8 x 10 4 } = 0.00424 M % ionized= 0.00424 0.10 x100% = 4.2 % less than 5%! approximation is OK poh = log [OH ] = log (0.00424)=2.37 ph=142.37=11.63 15
Weak bases React with water to produce OH Base + H 2 O BaseH + + OH hydrolysis reaction NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH (aq) Do not react completely Equilibrium constant K b 16
Aniline is a weak base. If a 0.95 M solution of aniline is 0.00202 % ionized, what is the K b value for aniline? 17
Determine the ph of a 4.0 x 10 3 M solution of methylamine (CH 3 NH 2 ). 18
19
20
Weak acids and the salts of their conjugate bases Acids H + + anion H+ anion If an acid is uncharged its conjugate base is negatively charged Conjugate bases of acids exist as ionic compounds aka salts Often from group I since all group I salts are soluble Salts of acids replace H + any cation + anion Na + anion K + anion Strong acids HCl HNO 3 H 2 SO 4 H + Cl H + NO 3 2H + SO 2 4 hydrochloric acid nitric acid sulfuric acid Weak acids HC 2 H 3 O 2 HF H + C 2 H 3 O 2 H + F acetic acid hydrofluoric acid Salts of Strong acids LiCl NaNO 3 K 2 SO 4 Li + Cl Na + NO 3 2K + SO 2 4 lithium chloride sodium nitrate potassium sulfate Salts of Weak acids Mg(C 2 H 3 O 2 ) 2 CsF Mg 2+ 2C 2 H 3 O 2 Cs + F magnesium acetate cesium fluoride Weak base compound with lone pairs often a N containing compound H N H Weak bases and the salts of their conjugate acids ammonia H H N R H If a base is uncharged its conjugate acid is positively charged Conjugate acids of bases exist as ionic compounds aka salts amines H H H N H Salt of its conjugate acid + anion H H and H N R H + anion When they act as bases gaining H + they become positively charged examples NH 4 Cl CH 3 NH 3 (ClO 4 ) H 21
Salts of weak acids/bases If soluble fully dissociate into ions Initial molarity calculated from chemical formula 0.33 M NaC 2 H 3 O 2 a solution 0.33 M in C 2 H 3 O 2 0.24 M Ca(C 2 H 3 O 2 ) 2 a solution 0.48 M in C 2 H 3 O 2 2 C 2 H 3 O 2 per 1 Ca(C 2 H 3 O 2 ) 2 0.24 M NH 4 Br a solution 0.24 M in NH 4 + Ions of group I Li +, Na +, K + = etc. have no effect on ph Ions of group II Mg 2+, Ca 2+, Ba 2+ = etc. have a very small acidic effect on ph Both are often treated as being spectator ions that do not affect ph if any other acid or base is present at a significant concentration. K a K b = K w For acetic acid the hydrolysis reaction is HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 2 (aq) + H 3 O + (aq) K a =1.76 x10 5 for acetate ion the hydrolysis reaction is C 2 H 3 O 2 (aq) + H 2 O HC 2 H 3 O 2 (aq) + OH (aq) K b =5.68 x10 10 notice if you add them the conjugate acid and base cancel overall reaction becomes 2H 2 O(l) H 3 O + (aq) + OH (aq) K w =? 10 14 Remember when reactions are added the overall K eq is the product of the K eq s K a K b = 1.76 x10 5 x 5.68 x10 10 = 10 14!!!!!!!!! 22
You will not have this half of the table for the exam You will not have this half of the table for the exam 23
ICE tables for salts of weak acids What is the ph of a 0.66 M solution of sodium acetate? C 2 H 3 O 2 (aq) + H 2 O HC 2 H 3 O 2 (aq) + OH (aq) Initial Change Equililibrium 0.66 0 0 x +x +x 0.66x +x +x By the usual approximation x = square root (0.66*5.68 x 10 10 ) = 1.936 x 10 5 [OH ] = 1.936 x 10 5 poh = 4.71 ph = 14 4.71 = 9.29 acetate ion K b = 10 14 K a (acetic acid) A basic solution b/c we added the conjugate base of acetic acid = 10 14 1.76 x10 5 = 5.68 x10 10 This problem is setup like other weak base problems, but you must recognize it as the conjugate base of a weak acid ICE tables for salts of weak bases What is the ph of a 0.66 M solution of ammonium chloride? NH 4 + (aq) + H 2 O NH 3 (aq) + H 3 O + (aq) initial change equil 0.33 0 0 x +x +x 0.33x +x +x By the usual approximation x = square root (0.33*5.56 x 10 10 ) = 1.35 x 10 5 [H 3 O + ] = 1.35 x 10 5 ph = 4.87 An acidic solution b/c we added the conjugate acid of ammonia Ammonium ion K a = 10 14 K b (ammonia) = 10 14 1.8 x10 5 = 5.56 x10 10 24
(so weak they do not affect ph) alcohols C 2 H 5 OH H 2 O weaker weak acids HCN Weak acids stronger weak acids HClO 2 Strong Acids HCl, HBr, HI, HNO 3, HClO 4, H 2 SO 4 Increasing Acid Strength Strong bases C 2 H 5 O OH Conjugate bases of weak acids stronger weak bases CN weaker weak bases ClO 2 Conjugate bases of Strong Acids (so weak they do not affect ph) Cl, Br, I, NO 3, ClO 4, HSO 4 Increasing Base Strength Determine the ph of a 0.246 M solution of methyl amine bromide (CH 3 NH 3 Br) 25
Determine the ph of a 0.468 M solution of sodium benzoate (NaC 7 H 5 O 2 ) 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 2 H 3 O 2 H + H + HC 2 H 3 O C 2 2 H 3 O 2 HC 2H 3 O 2 H + 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 26
Making a Buffer How buffers work against strong acid new HA H 2 O HA + H 3 O + HA (an acid) A (conjugate base) Added H 3 O + 27
new A How buffers work against strong base H 2 O HA A HA A + H 3O + Added HO pk a and pk b K a and K b values vary over a wide range Smaller than10 100 Larger than10 +100 Often shown in tables as pk a or pk b to avoid need for scientific notation pk a = log(k a ) pk b = log(k b ) Stronger acids have lower pk a values 28
HendersonHasselbach Equation Used to calculate ph of buffer solution Equation derived from the K a expression Input = initial concentrations of the weak acid and conjugate base Where: pk a = log(k a ) ph p K a B log A B = molarity or # of moles of the base A = molarity or # of moles of the acid 29
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 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 The other substance is its conjugate base It is a buffer so use the HHA equation Effectiveness of Buffers A buffer will be most effective against acids and bases when B/A = 1 equal amounts of acid and base Reasonably effective when 0.1 < B/A < 10 Buffers always have a larger capacity when B and A amounts are both higher 30
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) [B]/[A] = 1.0/0.85 = 1.17 pk a = log (4.6 x10 4 ) = 3.337 this buffer has same B/A ratio and therefore same ph as previous example 31
Selecting a conjugate acid/base pair for a buffer Buffers are most effective against acids and bases when B=A or B/A=1 This occurs when ph = pk a This is achieved by selecting an acid with a pk a close to desired ph This will make the B/A ratio close or equal to to 1.0 more effective buffer against acids and bases 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 32
Example 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 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 33
Example. Calculate the B/A ratio for a buffer based on chloroacetic acid and its conjugate base with a ph=3.0 34
Example: Calculate how many grams of sodium trichloroacetate must be combined with 1.0 gram of trichloroacetic acid in order to make a buffer with a ph of 1.0. 35
36