x x10. Hydromiun ion already in solution before acid added. NH 3 /NH4+ buffer solution

Transcription

1 10/15/01 Commonion effect In the last chapter, we calculated the [H 3 O ] of a M O as M. The percent dissociation for this solution would be: More Acid and Base Chemistry % [H 3 O ][ClO ] [O] ()(0.100 ) ()(0.100) Assume these s will be very small But what would happen if we place moles of O in a 1.000L container that already has 0.10es of the conjugate base NaClO. Will this change anything? I C E nitial hange O(aq) H O(l) H 3O (aq) ClO (aq) M quilibirum M % Hypoclorite ion already in solution before acid ed Note the change in percent dissociation. The percent dissociation decreased because there were already ClO ions in solution. The reaction did not proceed as far toward the reactants to reach equilibrium. This is the commonion effect. Common ions in solution decrease dissociation of weak acids and bases % Commonion effect What would happen if the solution had a ph of before ing M O? [H 3 O ][ClO ] [O] (0.100 )() Assume these s will be very small I C E nitial hange O(aq) H O(l) H 3O (aq) ClO (aq) M 0.10 quilibirum M (0.100)() Both products of dissociation of O, ClO and H 3 O, caused a decrease in the dissociation of the acid. This is the common ion effect Hydromiun ion already in solution before acid ed % How would ionization be affected if both H 3 O and ClO were 0.10 initially? Hint: think about Q a Buffers Buffers have the ability to stabilize ph, even with the ition of acids or bases. Buffers are composed of weak conjugate bases and acids in equilibrium. Plain ole water NH NH H ph = 7 ph = 0 /NH buffer solution NH NH ph = 9.5 NH Cl NH Cl NH NH 3 NH NH ph = 9.03 ΔpH = 7 s H 3O to the solution with its reaction with water. HO H 3 O Cl Through the reaction, 1 mol of H 3O is ed to the water. ΔpH = 0. reacts with to give more NH ions. H 3O can only be ed to the solution through the /NH equilibrium NH Cl Through the reaction, only mol of H 3O is ed to the water. Calculating ph of buffer solutions Calculate the ph of a buffer solution in a 1.0L container with 3.0 moles of and 5.0 moles of NH. K a for NH is [H3O ][] [NH ] ()(3.0) ()(3.0 ) ph = [H 3O ] = ( ) = 9.01 Henderson Hasslebalch Formula ph p K a NH (aq) H O(l) H 3O (aq) (aq) But instead of another I.C.E. table, we can use 3. ph in acid/base concentrations by dissociation negligible. [acid/base] initial [acid/base] equilibrium Let s do the same calculation above, but using HH ph=9.0 Because the [] [NH ] = 8., this is an 8.0 M buffer ing Strong Acid/Base to a Buffer Solution Let s take our first buffer problem from the last page: Calculate the ph of a buffer solution in a 1.0L container with 3.0 moles of and 5.0 moles of NH. K a for NH is What happens when we 0.50L of.0 to the solution First, we need to determine how ing changes the /NH buffer. To do this, we will use a change table. It is best to use moles not M in this table (you will see why later). Note that the (acid) will react with the (base) in the solution. ph p K The strong acid or base will always react completely. Remember, 100% dissociation. p = = ( ) = 9. a. ph ph=8.76 H (aq) (aq) NH (aq) : 1.e 3.es 5.es : 1.e 1.e 1.e e.es 6.es Since both and NH are in the same volume, we can use moles instead of concentration. If you divide both and by the same volume, the ratio stays the same, 1

2 10/15/01 Calculate the ph of a buffer solution in a 100.mL container with 0.10 of each component of a benzoic acid / sodium benzoate buffer. K a of benzoic acid is ph p K 0.10 a ph.19 ph What is the ph after ing 10.00mL of 0.50? OH (aq) (aq) (aq) : : Since there are still appreciable amounts of both and, it still a buffer solution. Thus, HH can be used Buffer Solution ph.19 ph OH What is the ph after ing a total of 0.00mL of 0.50 to the original (100.0mL 0.10 each) buffer solution? K a L L L 1L H Cl Cl OH (aq) (aq) (aq) : : Is this even a buffer solution any more? NO! Only! Then I cannot use HH! 0.00 A 0.167M A L L F (aq) H O(l) OH (aq) (aq) 0.167M M Weak Base Solution [OH ] ph 1 poh 1 ( 5.10 ) ph 8.7 What is the ph after ing a total of 30.00mL of 0.50 to the original (100.0mL 0.10 each) buffer solution? OH (aq) (aq) (aq) : : L poh ph OH M OH OH Strong Base Solution Which base should have a greater effect on ph? How a buffer reactions to ition of strong base or strong acid Assume we have a buffer with 1M total buffer strength. Let s start with 0.5 of each and. ing acid ing base ph Mol of conjugate base in 1mol total buffer ph p K a Note that our theoretical buffer has a K a of and thus a pk a of 7.

3 10/15/01 When does a buffer stop buffering? Titration problems (determining the ph at different points of a titration) are very similar to buffer questions. Use change tables, and see what s left. No longer buffering 1:1 base:acid Mol of conjugate base in 1mol total buffer A good rule of thumb: A buffer is relatively effective ± 1 ph point of the acid s pk a. that point the buffer is eghausted and can no longer buffer in that direction. 10:1 1:10 Determine the ph during of10.00ml of before the titration. What type of solution is this? Determine the ph during a titration after 1.50mL of has been ed to 10.00mL of OH (aq) H (aq) H O : mol mol : mol mol L mol H L mol 0.00M H ph (0.00M) ph. 66 H Cl H Cl Cl Cl H H Strong Acid Solution Cl Cl H Cl Cl H Strong Acid Solution ph (0.010) ph.000 Determine the ph during a titration after.00ml of has been ed to 10.00mL of Determine the ph during a titration after.50ml of has been ed to 10.00mL of OH (aq) H (aq) H O OH (aq) H (aq) H O : : mol mol Na mol Cl mol Cl Cl Cl : mol : mol mol mol mol Cl OH Cl Cl Cl ph 7.00 Neutral Solution mol OH 0.00 OH L L poh (0.00) 1.70 Strong Base Solution ph ph 1.30 AcidBase Titrations AcidBase Titrations Strong Acid/Strong Base Weak Acid/Strong Base ing Base to Acid Equivalence : The amount of base ed equals the amount of acid present before the titration. Equivalence : The amount of base ed equals the amount of acid present before the titration. acid/strong base titration the ph is 7. acid/strong base titration the ph is >7. ½ volume of equivalence point ph = pk a Eq. Pt. [] = [ ] (buffer zone) Vol of ed (ml) Vol of ed (ml) H Cl H Cl Cl Cl H H Cl Cl H Cl Cl H Cl Cl Cl Cl Cl OH Cl Cl Cl F F F F F F F Na OH F F F Titration Titration Equivalence Titration Titration Equivalence 3

4 10/15/01 AcidBase Titrations Strong Acid/Strong Base AcidBase Titrations Strong Acid/Weak Base Equivalence : The amount of acid ed equals the amount of base present before the titration. ing Acid to Base ½ volume of equivalence point ph = 1pK b [B]= [BH ] Equivalence : The amount of acid ed equals the amount of base present before the titration. Eq. Pt. (buffer zone) acid/strong base titration the ph is 7. acid/weak base titration the ph is <7. OH OH OH OH Cl OH OH Na Cl Cl Cl Cl Cl Cl Cl Na H Cl Cl Cl Cl NH Cl NH NH NH Cl Cl NH NH Cl Cl NH NH Cl H NH NH Cl Cl Titration Titration Equivalence Titration Titration Equivalence Strong Acid / Weak Base Titrations Determine the ph during of10.00ml of before the titration. K a = Determine the ph during a titration after 1.50mL of has been ed to 10.00mL of Determine the ph during a titration after.00ml of has been ed to 10.00mL of Determine the ph during a titration after.50ml of has been ed to 10.00mL of Polyprotic acid Titrations Since there are multiple protons to remove from a polyprotic acid, there are multiple equivalence points one for each proton. The the ph of the equivalence point of an amphoteric species (on that can either donate or accept an proton) pk pk n a 1 phof amphotericspecies a n 3 H 3 PO H PO HPO PO 3 pk a1 =.1 pk a = 7.0 pk a3 = 1.38 Br Solubility Product K sp In CHM151, we talked about ionic compounds that were insoluble in water. In reality, those compounds do dissolve, just a very. very small amount. For eample, based on our rules, AgBr is insoluble. How some very small amount does dissociate into ions. The represent this we have another K epression called K sp. Ag Since solids do not appear in any K epression K sp = [Ag ][Br ] and K sp = So, how do we determine the concentration of Ag and Br in the solution? The solubility of AgBr is represented by the equation: AgBr(s) Ag (aq) Br (aq) M Solubility Product K sp (more complicated eample) OH Mg OH How do we deal with sparingly soluble salts that do not have cations and anions in a 1:1 ratio? Mg(OH) (s) Mg (aq) OH (aq) [Mg ] = = 1.10 M [OH ] = = ()1.10 M =.910 M K sp = K sp = [Mg ][OH ] ()() = ()( ) = 3 = = 1.10 M [Ag ] = [Br ] = M This is called the molar solubility

5 10/15/01 Br Ag Solubility Product K sp solid AgBr Br Ag water Ag Br Br Br Ag Ag In all fours cases, the [Ag ] M and [Br ] is M Remember LeChatelier s Principle holds that the ition of a solid should not shift equilibrium. Does this hold true? YES! Assuming there is already some solid in the solution, the concentration of ions does not change. ly, as the water is ed, the concentration of ions is decreased. Based on LeChatelier s Principle the reaction should shift toward the product s side. Thus, in the end, more AgBr is dissolved and the concentration of ions is the same. Factors that effect solubility of sparingly soluble salts Commonions ph (acid/base reactions Common ions will decrease solubility of a sparingly soluble salt. Same as common ion effect. Acids will increase the solubility of a sparingly soluble salts with a basic ion (OH, F,etc.) through an acid/base reaction. OH Fe OH OH Cl Fe OH Fe Cl reestablished Cl Fe Cl Formation of Comple ions (Lewis acid/base reactions OH Cu OH a lot of Common ion of a sparingly soluble salt Let s say we have 0.50L of 0.5. Into this solution we drop in a large chunk of Mg(OH). Determine the concentrations of all ions in solution at equilibrium. Mg(OH) (s) Mg (aq) OH (aq) 0.5 K sp = OH 0.50 K sp = [Mg ][OH ] Mg(OH) = ()(0.50) = OH [Mg ] = = M [OH ] = 0.50 = 0.50 ()