CHM 112 Dr. Kevin Moore

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Transcription:

CHM 112 Dr. Kevin Moore

Reaction of an acid with a known concentration of base to determine the exact amount of the acid Requires that the equilibrium of the reaction be significantly to the right

Determination of the concentration of a solution Primary Standard highly pure compound (generally non-hygroscopic) used to determine the concentration of another compound Standard Solution solution of known concentration

Neutralization Constant is the K for the net ionic reaction Strong Acid /Strong Base 2 3 2 H O ( aq) OH ( aq) H O( l) K n 1 1 [ HO ][ OH ] K 3 w K n 1 10. 10 14 10. 10 14

Weak Acid/Strong Base HC H O ( aq) NaOH( aq) NaC H O ( aq) H O( l) 2 3 2 2 3 2 2 Net Ionic Equation HC H O ( aq) OH ( aq) C H O ( aq) H O( l) 2 3 2 2 3 2 2 K n can be determined from known equilbrium constants

Involves the dissociation of HC 2 H 3 O 2 and reaction with OH - HCHO( aq) HOl ( ) HO ( aq) CHO( aq) K a 18. 10 1 14 HO ( aq) OH 2HOl ( ) 10. 10 5 2 3 2 2 3 2 3 2 3 2 HC2H3O2( aq) OH ( aq) C2H 3O2 ( aq) H2O( l) 1 K w K K K n a w K 18. 10 9 n

Addition of a common ion shifts equilibrium Calculate the ph of a 0.10 M HC 2 H 3 O 2 solution which also contains 0.10 M NaC 2 H 3 O 2. HCHO( aq) HOl ( ) HO ( aq) CHO ( aq) K a 18. 10 5 2 3 2 2 3 2 3 2 0.10 0.00 0.10 -x +x +x K a [ x][ 010. x] [. 010 x] x H O K ph pk 3 a a

Calculate the ph of a solution of 0.15 M HF with 0.050 M NaF. HF( aq) H2O( l) H 3O ( aq) F ( aq) K a 35. 10 0.15 0.00 0.050 -x +x +x K a [ x][ 0. 050 x] [. 015 x] x H O 015. K 3 0050. 4 0050. x 015. a 00010. ph 300.

Solution which resists drastic changes to ph Weak Conjugate Acid/Base Pair NH 3 /NH 4 + HF/F - Biological systems H 2 CO 3 /HCO 3 -

Consider 1.0 L solution containing 0.10 M HC 2 H 3 O 2 and 0.10 M NaC 2 H 3 O 2. What is the ph after the addition of 0.010 moles of NaOH? HC H O ( aq) OH ( aq) C H O ( aq) H O( l) 2 3 2 2 3 2 2 0.10 0.01 0.10 -.01 -.01 +.010 0.09 0 0.11 HCHO( aq) HOl ( ) HO ( aq) CHO ( aq) K a 18. 10 0.09 [ x][ 011. x] 0.00 0.11 -x Ka +x +x [. 009 x] 5 2 3 2 2 3 2 3 2 x 15. 10 5 [ H3O ] ph 483.

What is the ph resulting from the addition of 0.020 moles of HCl to a 1.0L solution of 0.10 M HC 2 H 3 O 2 and 0.10 M NaC 2 H 3 O 2. CHO ( aq) HO ( aq) HCHO( aq) HOl ( ) 2 3 2 3 2 3 2 2 0.10 0.02 0.10 -.02 -.02 +.020 0.08 0 0.12 HCHO( aq) HOl ( ) HO ( aq) CHO ( aq) K a 18. 10 0.12 0.00 0.08 -x K [ x][ 008. x] a +x +x [. 012 x] 5 2 3 2 2 3 2 3 2 x 27. 10 5 [ H3O ] ph 457.

Measure of the amount of acid/base that a buffer solution can absorb depends on moles of acid/base present

Simple method of determining ph of buffer Base ph pka log [ ] [ Acid] Acid poh pkb log [ ] [ Base]

[Base/Acid]=1, log = 0 and ph=pka If HA has dissociated 99%, what would [Base/Acid] be? HA( aq) H O( l) H O ( aq) A ( aq) 2 3 0.10 0 0-0.099 0.099 0.099 0.001 0.099 0.099 A ph pka log [ ] ph pka [ HA] 0 log. 099 200. 0. 001 0099 log. 0001.

Capacity is ±2.0 ph units Rate of consumption depends on moles What is the ph of a buffer containing 0.20 M NaHCO 3 and 0.10 M Na 2 CO 3? (K a2 =5.6 x 10-11 ) 2 HCO ( aq) H O( l) H O ( aq) CO ( aq) 3 2 3 3 ph ph 2 CO3 010 pka2 log [ ] 1 10. 25 log. [ HCO ] 020. 995. 3

Equivalence point ph at which equivalent amounts of acid and base have been added EndPoint Amount of titrant needed to reach the equivalence point 4 important areas of the curve Pure Acid Mixture of Acid/Base Salt Pure Base

ml x M = mmol easy to work with if solutions are dilute How many mmol of acid are present in 25.0 ml of 0.20 M solution? 020. M 250. ml 50. mmol acid

25.0 ml of 0.25M HC 2 H 3 O 2 is titrated with 0.30 M NaOH 1) initial ph regular weak acid ph 2) ph after 5.0 ml of base Buffer 3) volume of base and ph at eq pt Salt of conj. base and neutral ion 4) ph after 1.0 ml of base added beyond eq pt pure base NaOH( aq) HC2H3O2( aq) NaC 2H3O2( aq) H2O( l) mmol HC H O 025. M 250. ml 2 3 2 mmol HC H O 2 3 2 625. mmol

1) initial ph HCHO( aq) HOl ( ) HO ( aq) CHO ( aq) K a 18. 10 5 2 3 2 2 3 2 3 2 0.25 0.00 0.00 -x +x +x K a [ x][ x] [. 025 x] x H O 5 3 18. 10 025. 21. 10 3 ph 267.

2) ph after 5.0 ml of NaOH has been added mmol OH 030. M 50. ml 15. mmol HC H O ( aq) NaOH( aq) NaC H O ( aq) H O( l) 2 3 2 2 3 2 2 6.25 mmol 1.50 mmol 0.00-1.50-1.50 +1.50 4.75 mmol 0 1.50 Buffer Solution (significant quantities of conjugate acid/base pair) ph ph 150 474. log [. ] [. 475] 424.

3) Volume of base at equivalence point mmol OH 625. 625. mmol V 208. ml 030. M mmol HC H O 2 3 2

ph at Equivalence point HC H O ( aq) NaOH( aq) NaC H O ( aq) H O( l) 2 3 2 2 3 2 2 6.25 mmol 6.25 mmol 0.00-6.25-6.25 +6.25 0.00 0.00 6.25 mmol [ CHO ] 2 3 2 625. mmol 458. ml 014. Dissociation of conjugate base (K b/ ) K K ' w. b Ka 10 10 18. 10 14 5 56. 10 10

ph at Equivalence Point C H O ( aq) H O l HC H O ( aq) OH ( aq) 2 3 2 2 2 3 2 0.14 0.00 0.00 -x +x +x K x / b x 014. 2 x [ OH ] 014. ( 56. 10 ) 89. 10 10 6 NaC 2 H 3 O 2 is a basic salt poh ph 505. 895.

4) ph at 1.00 ml of base beyond eq pt ml Base = 21.8 218. ml 0. 30 M 654. mmol 654. 625. 029. mmol 029. mmol [ OH ] 62. 10 3 468. ml poh 221. ph 1179.

11.79 8.95 ph 4.24 2.67 5.0 20.8 21.8 ml of Base

Verify inflection Calculate ph at 20.0 ml 2 3 2 2 3 2 2 6.25 mmol 6.00 mmol 0.00-6.00-6.00 +6.00 0.25 mmol 0 6.00 ph ph mmol OH 030. M 200. ml 600. mmol HC H O ( aq) NaOH( aq) NaC H O ( aq) H O( l) Buffer Solution (significant quantities of conjugate acid/base pair) 600 474. log [. ] [. 025] 612.

11.79 8.95 6.12 4.24 2.67 5.0 20.0 20.8 21.8 ml of Base

Complete data set ph 14.00 12.00 10.00 8.00 6.00 ph 4.00 2.00 0.00 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0

What happens when titrating a diprotic acid? Protons react in succession Two buffering solutions Two equivalence points

ph=pk a2 K b/ = K w K a2 ph=pk a1 ph=pk a1 +pk a2 2 ml of Base

HIn (aq) H + (aq) + In - (aq) [HIn] [In - ] [HIn] [In - ] 10 10 Color of acid (HIn) predominates Color of conjugate base (In - ) predominates

Insoluble solids Some level of solubility generally undetectable to eye Governed by equilibrium 2 Mg( OH) ( s) Mg ( aq) 2OH ( aq) 2 K [ Mg ][ OH ] sp 2 2

CaCO ( s) Ca ( aq) CO ( aq) 3 2 2 3 2 Ksp [ Ca ][ CO3 ] 50. 10 2 9

What is the solubility Product Constant if the maximum solubility of Mg(OH) 2 is 1.7 x 10-4 M. 2 Mg( OH) ( s) Mg ( aq) 2OH ( aq) 2 K [ Mg ][ OH ] sp 2 2 4 4 2 K [. 17 10 ][. 34 10 ] sp K sp 20. 10 11

# of moles of solute dissolved per Liter of solution What is the molar solubility of Ag 2 CrO 4? K sp = 1.7 x 10-11? 2 Ag CrO () s 2 Ag ( aq) CrO ( aq) 2 4 0.00 0.00 4 +2x +x 2 2 K [ Ag ] [ CrO ] sp 17. 10 [ 2x] [ x] 4x 11 2 3 4 x 16. 10 4

What is the molar solubility of AgCl? (K sp =1.8x10-10 ) AgCl( s) Ag ( aq) Cl ( aq) 0.00 0.00 +x +x K [ Ag ][ Cl ] sp 18. 10 10 [ x][ x] x 2 x 13. 10 5

Presence of a common ion decreases the solubility of an insoluble solid What is the molar solubility of AgCl in 0.10 M KCl? AgCl( s) Ag ( aq) Cl ( aq) K [ Ag ][ Cl ] sp 0.00 0.10 +x +x 18. 10 10 [ x][ 010. x] 010. x x 18. 10 9

Can insoluble salts be made more soluble? If ion is present which removes one of the equilibrium ions Metals ions in presence of Lewis Bases Conjugate Base in presence of Strong Acid CO 3-2, PO 4-3, SO 3-2, OH - 2 FeCO () s Fe ( aq) CO ( aq) K 35. 10 3 2 11 3 sp 2 H O ( aq) CO ( aq) HCO ( aq) H O( l) 3 3 K K K 63. 10 1 sp n 3 2 K n K K b w 18. 10 10. 10 4 14 18. 10 10

Tooth Enamel: Ca PO OH( s) Ca PO ( aq) OH ( aq) 5 4 3 5 4 3 Ca PO OH() s H O ( aq) Ca PO ( aq) 2H O() l 5 4 3 3 5 4 3 2

Metal Ions act as Lewis Acids Bond with Lewis Base to make complex ions Lewis Bases: lone pair donors H 2 O, OH -, NH 3, Cl- Ag ( aq) 2NH3( aq) Ag( NH3) 2( aq) K f 17. 10 7

Substances which react with both acid and base Al +3 compounds tend to be amphoteric 3 Al O ( s) 6H O ( aq) 2Al ( aq) 9H O( l) 2 3 3 Al O ( s) 2OH ( aq) 3H O l 2Al( OH) ( aq) 2 3 2 4 2

Al(OH) 3 4 10 ph Solubility

Determine if ion concentration is such that a precipitate will appear IP > K sp ===> PPT! IP = K sp ===> saturated solution IP < K sp ===> unsaturated solution Use K sp expression to determine IP

If 150. ml of 0.10 M Pb(NO 3 ) 2 is mixed with 100. ml of 0.20 M NaCl, will solution ppt (K sp (PbCl 2 ) = 1.2 x 10-5 )? Pb( NO ) ( aq) 2NaCl( aq) PbCl ( s) 2NaNO ( aq) 3 2 2 3 2 PbCl ( s) Pb ( aq) 2Cl ( aq) 2 2 0015. [ Pb ] 010. M( 0150. L) 0060. 0250. 0020. [ Cl ] 020. M( 0100. L) 0. 080 0250. IP [ Pb ][ Cl ] [ 0060. ][ 0080. ] IP 2 2 2 4 38. 10 IP>K sp PPT will be present

If 50.0 ml of 3.0 x 10-3 M BaCl 2 is mixed with 75.0 ml of 2.0 x 10-3 M Na 2 CO 3, will a ppt form (K sp (BaCO 3 ) = 2.6 x 10-9 )? BaCl ( aq) Na CO ( aq) 2NaCl( aq) BaCO ( s) 2 2 3 3 2 2 BaCO () s Ba ( aq) CO ( aq) 3 2 15. 10 [ Ba ] 0. 0030M( 0. 0500 L) 0125. 2 15. 10 [ CO3 ] 0. 0020M( 0. 0750 L) 0125. 2 2 IP [ Ba ][ CO ] [ 00012. ][ 00012. ] IP 3 3 4 4 00012. 00012. 6 14. 10 IP>K sp PPT will be present

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