# Buffer Solutions. Buffer Solutions

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1 Buffer Solutions ph of solution adding 0.10 M HCl to 100 ml water HCl added ph 0 ml ml ml ml ml ml ml of 0.10 M HCl added Buffer Solutions A buffer helps a solution maintain its ph when acid or base is added A buffer must contain two components to work a weak acid that reacts with added base a weak base that reacts with added acid Buffers usually contain approximately equal amounts of a weak acid and its conjugate base 1

2 Buffer Solutions Solution that is M CH 3 COOH (acetic acid) and M NaCH 3 COO (sodium acetate) Find ph of buffer solution: CH 3 COOH(aq) + H 2 O CH 3 COO - (aq) + H 3 O + (aq) [CH 3 COOH] [CH 3 COO - ] [H 3 O + ] initial x x x equil x x x Buffer Solutions Find ph of buffer solution: CH 3 COOH(aq) + H 2 O CH 3 COO - (aq) + H 3 O + (aq) K a = [CH 3COO - ][H 3 O + ] [CH 3 COOH] = ( x)x ( x) = 1.8 x 10-5 x = 1.80 x 10-5 M assume x is negligible compared to.100 M ph =

3 Buffer Solutions Add 5 ml.10 M HCl Find ph of resulting solution Assume all acid added reacts with acetate ion to form acetic acid (remember that acids react with bases) CH 3 COOH(aq) + H 2 O CH 3 COO - (aq) + H 3 O + (aq) [H 3 O + ] added = (5 ml)(.10 M)/(105 ml) = 4.76x10-3 M [CH 3 COOH] = (0.100)(100/105) = M [CH 3 COO - ] = (0.100)(100/105) = M (100/105)=dilution factor for addition of HCl Buffer Solutions Now let solution come to equilibrium [CH 3 COOH] [CH 3 COO - ] [H 3 O + ] initial x x x equil.100 x x x K a = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] = ( x)x ( x) = 1.8 x 10-5 x = 2.00 x 10-5 M ph =

4 Buffer Solutions ph of buffered solution adding 0.10 M HCl to 100 ml soln HCl added ph 0 ml ml ml ml ml ml Buffered solution ml of 0.10 M HCl added Buffer Solutions Henderson-Hasselbach Equation Allows calculation of ph of a buffer if concentrations of conjugate acid and conjugate base are known HA(aq) + H 2 O H 3 O + (aq) + A - (aq) K a = [H 3O + ][A - ] [HA] [H 3 O + ] = K [HA] a [A - ] 4

5 Buffer Solutions Take -log of both sides log[h 3 O + ] = -log K a[ha] [A - ] = - log K a [A - ] ( ) -log [HA] -log(k a ) = pk a log [HA] = log [A- ] [A ] [HA] ph = pk a + log [A- ] [HA] Henderson-Hasselbach Eqn Buffer Solutions Using the Henderson-Hasselbach Eqn, we can: Determine ph of a solution Determine ratio of conjugate base to conjugate acid to achieve specific ph ph = pk a + log [A- ] [HA] 5

6 Buffer Solutions Let s go back to problem of adding HCl to buffer solution: We can use H-H eqn. to make the calculations much easier [CH 3 COOH] = [HCl] added [CH 3 COO - ] = [HCl] added ph = pk a + log [A- ] [HA] Buffer Solutions V HCl [HCl] [CH 3 COOH] [CH 3 COO - ] ph 5mL (.1)(5mL)/105mL = M =.100 M =.090 M mL (.1)(10)/ = M =.100 M =.082 M mL (.1)(25)/ =.0200 M =.100 M =.060 M mL (.1)(50)/ =.0333 M =.100 M =.034 M

7 Buffer Solutions Buffer Capacity the amount of acid or base that can be added to a buffer without the ph significantly changing Suppose we acid to a buffer solution: The acid will react with the conjugate base until it is depleted Past this point, the solution behaves as if no buffer were present Acid-Base Titrations A titration is a method used to determine the concentration of an unknown species Add a measured amount of a known reactant Determine when the reaction has gone to completion [unknown] + [known] products At the equivalence point moles unknown = moles known added C unknown V unknown = C known V known 7

8 Acid-Base Titrations Titrate ml unknown HCl soln. with M NaOH equivalence point ml NaOH added Acid-Base Titrations At equivalence point, V NaOH = ml mol(naoh) = (.2137 M)( L) = x 10-3 mol mol(hcl) = x 10-3 mol (mol known = mol unknown) [HCl] = (5.120x10-3 mol)/( L) = M equivalence point ml NaOH added 8

9 Indicators An indicator is a chemical species that changes color depending on the ph of the solution An indicator is a conjugate acid-conjugate base pair in which the acid and base forms of the compound have different colors HIn(aq) + H 2 O In - (aq) + H 3 O + (aq) color 1 color 2 Indicators are used to determine the endpoint of a titration Indicators The pk a of the indicator determines the ph range over which the color changes [HIn]/[In - ] 10 acid color [HIn]/[In - ] 0.1 base color [HIn]/[In - ] 1 intermediate color Remember: ph = pk a + log{[in - ]/[HIn]} If [HIn]/[In - ] = 1, log{[hin]/[in - ]} = 0 ph = pk a at point when indicator is changing color 9

10 Indicators Indicator pk a ph range color change Methyl orange red to yellow Bromophenol blue yellow to blue Methyl red red to yellow Bromothymol blue yellow to blue Phenol red yellow to red Phenolphthalein clear to pink Indicators Figure 17.5: ph curve for titration of M HCl with M NaOH change after endpoint change around endpoint change before endpoint 10

11 Indicators Titration of weak acid with strong base HA(aq) + OH - (aq) A - (aq) + H 2 O At equivalence point A - (aq) + H 2 O HA(aq) + OH - (aq) the solution is basic because conjugate base of weak acid reacts with water to form OH - (aq) Indicators Titrate ml M formic acid (HCOOH) with M NaOH K a = 1.8 x 10-4 Find ph at equivalence point and select appropriate indicator At equivalence point, mol(hcooh) = mol(oh - ) mol(fa) = (0.100 M fa)( L) = 2.5 x 10-3 mol fa = formic acid V NaOH added = (2.5 x 10-3 mol)/(0.100 M) = 25.0 ml V total = 50.0 ml 11

12 Indicators Assume HCOOH + OH - reaction goes to completion: [HCOO - ] = (2.5 x 10-3 mol)/( L) = M Determine K eq for reaction of formate ion: HCOO - (aq) + H 2 O HCOOH(aq) + OH - (aq) K eq = [HCOOH][OH- ] [HCOO ] = K w K a = 6.67 x Indicators K eq = [HCOOH][OH- ] [HCOO ] = K w K a = 6.67 x [HCOO - ] [HCOOH] [OH - ] initial x x x equil.0500 x x x K eq = [HCOOH][OH- ] [HCOO - ] = x x = 6.67 x

13 Indicators K eq = [HCOOH][OH- ] [HCOO - ] = x x = 6.67 x x = 1.83 x 10-6 M = [OH - ] poh = -log(1.83 x 10-6 ) =5.74 ph = = 8.26 Phenol red ( ) or phenolphthalein ( ) would be appropriate indicators Indicators Titration of weak base with strong acid B(aq) + H 3 O + (aq) BH + (aq) + H 2 O At equivalence point BH + (aq) + H 2 O B(aq) + H 3 O + (aq) the solution is acidic because conjugate acid of weak base reacts with water to form H 3 O + (aq) 13

14 Indicators Figure 17.8: titration of M NH 3 with M HCl Acid Rain Carbon dioxide in the air is in equilibrium with H 2 O in atmospheric water droplets (clouds & fog): CO 2 (aq) + H 2 O H 2 CO 3 (aq) carbonic acid K a = 4.2 x 10-7 H 2 CO 3 (aq) + H 2 O H 3 O + (aq) + HCO 3- (aq) Natural rain water has ph =

15 Acid Rain Emitted pollutants can form additional acid sources in clouds: NO 2 : 2 NO 2 (aq) + H 2 O HNO 3 (aq) + HNO 2 (aq) nitric acid nitrous acid strong K a = 4.5 x 10-4 Acid Rain Emitted pollutants can form additional acid sources in clouds: SO 2 : SO 2 (aq) + H 2 O H 2 SO 3 (aq) sulfurous acid K a = 1.2 x SO 2 (g) + O 2 (g) 2 SO 3 (g) SO 3 (aq) + H 2 O H 2 SO 4 (aq) sulfuric acid strong 15

16 Acid Rain Solubility Products Many salts are only slightly soluble The solubility product is a measure of the concentration of ions in a solution saturated with the salt MA(s) M + (aq) + A - (aq) K sp = [M + ][A - ] Examples AgCl(s) Ag + (aq) + Cl - (aq) K sp =[Ag + ][Cl - ]=1.8x10-10 PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp =[Pb 2+ ][Cl - ] 2 =1.7x10-5 AuBr 3 (s) Au 3+ (aq) + 3 Br - (aq) K sp =[Au 3+ ][Br - ] 3 =4.0x

17 Solubility Products Knowing the K sp, we can calculate the concentration of ions in solution Examples AgCl(s) Ag + (aq) + Cl - (aq) K sp =[Ag + ][Cl - ]=1.8x x x x x 2 = 1.8 x x = 1.3 x 10-5 M = [Ag + ] = [Cl - ] Solubility Products Examples PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp =[Pb 2+ ][Cl - ] 2 =1.7x10-5 -x x 2x x(2x) 2 = 1.7 x x 3 = 1.7 x 10-5 x = 1.6 x 10-2 M [Pb 2+ ] = 1.6 x 10-2 M [Cl - ] = 3.2 x 10-2 M 17

18 Solubility Products Examples AuBr 3 (s) Au 3+ (aq) + 3 Br - (aq) K sp =[Au 3+ ][Br - ] 3 =4.0x x x 3x x(3x) 3 = 4.0 x x 4 = 4.0 x x = 6.2 x M [Au 3+ ] = 6.2 x M [Br - ] = 1.9 x 10-9 M Solubility Products Examples Common ion effect How much PbI 2 will dissolve in a M solution of NaI? PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = 8.7 x x x 2x x(2x+.0100) 2 = 8.7 x 10-9 x(4x x + 1.0x10-4 ) = 8.7 x x x x10-4 x 8.7x10-9 = 0 x = 8.6 x 10-5 M vs 1.3 x 10-3 M if no I - (aq) were present initially 18

19 Factors Affecting Solubility Salts that are slightly soluble in water can be much more soluble in acid if one or both of its ions are moderately basic: CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) K sp = 8.7 x 10-9 But CO 3 2- (aq) is the conjugate base of HCO 3- (aq) CO 3 2- (aq) + H 3 O + (aq) HCO 3- (aq) + H 2 O K b = 2.1x10-4 HCO 3- (aq) + H 3 O + (aq) H 2 CO 3 (aq) + H 2 O K b = 2.4x10-8 H 2 CO 3 (aq) H 2 O + CO 2 (g) K eq 10 5 Works for carbonates, some sulfides, phosphates, etc. (species that behave as bases [no too weak]) Precipitation Define ion quotient, Q, as: M x A y (s) x M y+ (aq) + y A x- (aq) Q = [M y+ ] x [A x- ] y A precipitate will form only when Q exceeds K sp Q < K sp : solution is unsaturated no precipitate Q > K sp : solution is saturated precipitate forms Q = K sp : solution at saturation point 19

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