Chap16-Part3

# Chap16-Part3 -...

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Unformatted text preview: EXAM‐3: APRIL 15th 2011 (Friday)  Chapters: 15, 16 and 17  Weak Acids  Most acids only par/ally dissociate in water:   HA(aq)  +  H2O(l)  ↔  H3O+(aq)  + A‐(aq)   HA(aq)  ↔  H+(aq)  + A‐(aq)  The acid dissocia/on constant, Ka, describes the extent of  protonated and deprotonated species at equilibrium:  Larger values of Ka indicate stronger acids (more H+ generated).  When a compound is composed en/rely of C, H and O atoms, the  hydrogen atoms which dissociate are almost always bonded  to O rather than C atoms.  CalculaHng Ka and the Percent IonizaHon from pH  1) Make a table with the known informa/on.  2) Calculate [H+] from pH.  3) Write the concentra/on changes.  4) Calculate the equilibrium concentra/ons.  5) Find the percent ioniza/on.  6) Calculate Ka.  Example:  0.085 M phenylace/c acid (HC8H7O2) with pH = 2.68   HC8H7O2(aq)  ↔  H+(aq)  + C8H7O2‐(aq)  1) Make a table with the known informa/on.  HC8H7O2 (M) initial change final H+ (M) C8H7O2- (M) 0.085 0 0 2) Calculate [H+] from pH.   HC8H7O2 (M) initial H+ (M) C8H7O2- (M) 0.085 0 0 change final 2.1x10-3 3) Write the concentra/on changes.  HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 2.1x10-3 4) Calculate the equilibrium concentra/ons.  HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 5) Find the percent ioniza/on.   HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 PS: Generally, percent Ioniza/on for an acid = [H+]equilib X 100/ [HA]ini/al  6) Calculate Ka.  HC8H7O2 (M) H+ (M) C8H7O2- (M) initial 0.085 0 0 change -2.1x10-3 +2.1x10-3 +2.1x10-3 final 0.083 2.1x10-3 2.1x10-3 CalculaHng pH from Ka  1) Make a table with the known informa/on.  2) Write the changes in concentra/ons as a func/on of   an unknown variable (e.g., “x”).  3) Calculate the equilibrium concentra/ons as a func/on of x.  4) Write an algebraic expression for Ka as a func/on of x.  5) Find the value of x.  6) Calculate the equilibrium concentra/ons.  7) Calculate pH from [H+].  Example:  The ac/ve ingredient in aspirin is acetylsalicylicacid (HC9H7O4).   Aspirin is a monopro/c acid with Ka = 3.3x10‐4 at 25oC.  Find the pH of a  solu/on made by dissolving two 500 mg aspirin tablets in 250 mL water.         HC9H7O4(aq)  ↔  H+(aq)  + C9H7O4‐(aq)  ionizable group  1) Make a table with the known informa/on.  HC9H7O4 (M) initial change final H+ (M) C9H7O4- (M) 0.022 0 0 2) Write the changes in concentra/ons as a func/on of   an unknown variable (e.g., “x”).  HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 3) Calculate the equilibrium concentra/ons as a func/on of x.  HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 0.022 - x x x 4) Write an algebraic expression for Ka as a func/on of x.  HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -x +x +x final 0.022 - x x x 5) Find the value of x.  6) Calculate the equilibrium concentra/ons.  HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -2.5x10-3 +2.5x10-3 +2.5x10-3 final 0.020 +2.5x10-3 +2.5x10-3 7) Calculate pH from [H+].  HC9H7O4 (M) H+ (M) C9H7O4- (M) initial 0.022 0 0 change -2.5x10-3 +2.5x10-3 +2.5x10-3 final 0.020 +2.5x10-3 +2.5x10-3 Strong acids release much more  H+ into aqueous solu/on than  weak acids.  The led ﬂask contains 1 M   HC2H3O2 (a weak acid).  The right ﬂask contains 1 M  HCl (a strong acid).  The balloons both contain  the same amount of magnesium.  When equilibrium is reached,  the strong acid on the right  releases much more H+ than  the weak acid on the led.  The released H+ converts  to H2 gas, and the balloon  on the right is more highly  inﬂated.  The percent ioniza/on and the pH depend on the acid concentra/on  ace/c acid  CalculaHng pH from Ka  1) Make a table with the known informa/on.  2) Write the changes in concentra/ons as a func/on of   an unknown variable (e.g., “x”).  3) Calculate the equilibrium concentra/ons as a func/on of x.  4) Write an algebraic expression for Ka as a func/on of x.  5) Find the value of x.  6) Calculate the equilibrium concentra/ons.  7) Calculate pH from [H+].  Calculate the percent ioniza/on and the pH for   (a)  0.100 M HF and (b) 0.00100 M HF.    The value of  Ka is 6.8 x 10‐4.  1a) Make a table with the known informa/on.  initial change final HF (M) 0.100 H+ (M) 0 F- (M) 0 2a) Write the changes in concentra/ons as a func/on of an unknown  variable (e.g., “x”).  HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 3a) Calculate the equilibrium concentra/ons as a func/on of x.  HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 0.100 - x x x 4a) Write an algebraic expression for Ka as a func/on of x.  HF (M) H+ (M) F- (M) initial 0.100 0 0 change -x +x +x final 0.100 - x x x 5a) Find the value of x.  6a) Calculate the equilibrium concentra/ons.  HF (M) H+ (M) F- (M) initial 0.100 0 0 change -7.9x10-3 +7.9x10-3 +7.9x10-3 final 0.092 +7.9x10-3 +7.9x10-3 7a) Calculate pH from [H+].  HF(M) H+ (M) F- (M) initial 0.100 0 0 change -7.9x10-3 +7.9x10-3 +7.9x10-3 final 0.092 +7.9x10-3 +7.9x10-3 1b) Make a table with the known informa/on.  HF (M) initial change final H+ (M) F- (M) 0.00100 0 0 2b) Write the changes in concentra/ons as a func/on of   an unknown variable (e.g., “x”).  HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 3b) Calculate the equilibrium concentra/ons as a func/on of x.  HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 0.00100 - x x x 4b) Write an algebraic expression for Ka as a func/on of x.  HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -x +x +x final 0.00100 - x x x 5b) Find the value of x.  6b) Calculate the equilibrium concentra/ons.  HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -5.5x10-4 +5.5x10-4 +5.5x10-4 final 0.00045 +5.5x10-4 +5.5x10-4 7b) Calculate pH from [H+].  HF (M) H+ (M) F- (M) initial 0.00100 0 0 change -5.5x10-4 +5.5x10-4 +5.5x10-4 final 0.00045 +5.5x10-4 +5.5x10-4 The percent ioniza/on and the pH depend on the acid concentra/on  For higher acid concentra/ons, the percent ioniza/on decreases,  but the concentra/on of [H+] increases and the pH decreases.  (a)  Which is the strongest acid?    (a)  Which is the strongest acid?    HY isthe strongest acid because it is completely dissociated.  (b) Which has the smallest Ka?    (b) Which has the smallest Ka?        HX  (c) Which solu/on has the highest pH?  (c) Which solu/on has the highest pH?  The solu/on of HX has the highest pH because it has the  lowest concentra/on of H+.   PolyproHc Acids  Polypro/c acids have more than one ionizable proton.  PolyproHc Acids  It is always easier to remove the ﬁrst proton as compared to the second.  Note that Ka1 > Ka2 > Ka3.  ...
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## This note was uploaded on 09/13/2011 for the course CHEM 102 taught by Professor Todd during the Spring '08 term at UNC.

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