ACIDS AND BASES REVISE STRONG ACIDS AND BASES. LiOH Li + + OH - IMPOSSIBLE!!!!!!

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1 ACIDS AND BASES REVISE Brøsted sted-lowry acids ad bases Amphoteric substaces Cojugate acid base pairs Neutralisatio Neutral: ph 7 ([H + ] [OH - ]) Acidic: ph < 7 ([H + ] > [OH - ]) Basic: ph > 7 ([H + ] < [OH - ]) ph -log [H + ] poh -log [OH - ] ph + poh 14 at 5 o C K w [OH - ][H + ] K w 14.0 at 5 o C K w K a K b STRONG ACIDS AND BASES Strog acids ad bases react early completely to produce H + ad OH - equilibrium costats are large Calculate the ph of 0.1M LiOH. e.g.: HCl H + + Cl - + [H ][Cl ] K [HCl] Commo strog bases: LiOH, NaOH, KOH, RbOH, CsOH, R 4 NOH Complete dissociatio: large small I fact, we assume the reactio goes to completio: HCl H + + Cl - Commo strog acids: HCl, HBr, HI, H SO 4, HNO 3, HClO 4 (Why is HF ot a strog acid?) Start: Complete rx: LiOH Li + + OH - Problem: What is the ph of 1x10-8 M KOH? As before: poh -log (1x10-8 ) 8 ph Sice the cocetratio of KOH is so low (1x10-8 M), we eed to take the ioisatio of water ito accout. I pure water [OH - ] 1x10-7 M, which is greater tha the cocetratio of OH - from KOH. We do this by systematic treatmet of equilibrium. Charge balace: [K + ] + [H + ] [OH - ] BUT ph 6 acidic coditios ad KOH is a strog base Mass balace: Equilibria: [K + ] 1x10-8 M [H + ][OH - ] K w 1x10-14 M IMPOSSIBLE!!!!!! 3 equatios + 3 ukows solve simultaeously Fid: ph 7.0 Hit: You ed up with a quadratic equatio which you solve usig the formula. 1

2 Also ote that: Oly pure water produces 1x10-7 M H + ad OH -. If there is say 1x10-4 M HBr i solutio, ph 4 ad [OH - ] 1x10-10 M But the oly source of OH - is from the dissociatio of water. if water produces 1x10-10 M OH - it ca oly produce 1x10-10 M H + due to the dissociatio of water. ph i this case is due maily to the dissociatio of HBr ad ot the dissociatio of water. It is thus importat to look at the cocetratio of acid ad bases preset. Some guidelies regardig the cocetratios of acids ad bases: 1) Whe coc > 1x10-6 M calculate ph as usual ) Whe coc < 1x10-8 M ph 7 (there is ot eough acid or base to affect the ph of water) 3) Whe coc 1x x10-6 M Effect of water ioisatio ad added acid ad bases are comparable, thus: use the systematic treatmet of equilibrium approach. Weak acids ad bases react oly partially to produce H + ad OH - equilibrium costats are small H + + A - + [H ] K a WEAK ACIDS AND BASES Partial dissociatio small large Acid dissociatio costat Commo weak acids: carboxylic acids (e.g. acetic acid CH 3 COOH) ammoium ios (e.g. RNH 3+, R NH +, R 3 NH + ) + H O H 3 O + + A - + [H3O ] Ka Commo weak bases: carboxylate aios (e.g. acetate CH 3 COO - ) amies (e.g. RNH, R NH, R 3 N) Base hydrolysis: B + H O BH + + OH - + [BH ][OH ] K b [B] base hydrolysis costat/ base associatio costat NOTE: pk a -log K a pk b -log K b Weak base partial dissociatio K b small As K icreases, its p-fuctio decreases ad vice versa. Problem: Fid the ph of a solutio of formic acid give that the formal cocetratio is M ad K a 1.80x10-4. Sice [HCOOH] > 1x10-6, we ca calculate ph as usual Start: Equilibrium: HCOOH H + + HCOO - FRACTION OF DISSOCIATION, α Fractio of acid i the form A - α + + [A ] α F ( F ) For the above problem: [HCOO α - ] F Acid is..% dissociated at M formal cocetratio Weak electrolytes dissociate more as they are diluted. ph 1.7

3 B + H O BH + + OH - Charge balace: [BH + ] [OH - ] Mass balace: F [B] + [BH + ] Equilibria: WEAK BASE EQUILIBRIA Let [BH + ] [OH - ] x x. x x Kb [B] F x K b [BH + ] α F + [BH ][OH ] [B] FRACTION OF ASSOCIATION CONJUGATE ACIDS AND BASES Relatioship betwee K a ad K b for a cojugate acidbase pair: K a.k b K w 1x10-14 at 5 o C If K a is very large (strog acid) The K b must be very small (weak cojugate base) Ad vice versa If K a is very small, say 1x10-6 (weak acid) Base so weak it is ot a base at all i water The K b must be small, 1x10-8 (weak cojugate base) Greater acid stregth, weaker cojugate base stregth, ad vice versa. Problem: Calculate the ph of 0.1 M NH 3, give that pk a 9.44 for NH 4+. BUFFERS Mixture of a weak acid ad its cojugate base. Buffer solutio resists chage i ph whe acids or bases are added or whe dilutio occurs. Mix: A moles of weak acid + B moles of cojugate base Fid: moles of acid remais close to A, ad moles of base remais close to B Very little reactio ph 11.1 H + + A - Le Chatelier s priciple HENDERSON-SSELBALCH SSELBALCH EQUATION For acids: ph pka For bases: [B] ph pka + [BH ] K b B + H O BH + + OH - K base acid a acid base Whe [A - ], ph pk a pk a applies to this acid Derivatio: H + + A - K a + [H ] + [H ] -logka log + log[h ] log pk a ph log ph pka 3

4 ? Why does a buffer resist chage i ph whe small amouts of strog acid or bases is added? The acid or base is cosumed by A - or respectively A buffer has a maximum capacity to resist chage to ph. Buffer capacity, β: Measure of how well solutio resists chage i ph whe strog acid/base is added. dcb dc β a dph dph H + + A - A buffer is most effective i resistig chages i ph whe: ph pk a i.e.: [A - ] Choose buffer whose pk a is as close as possible to the desired ph. pk a ± 1 ph uit Larger β more resistace to ph chage Problem: Calculate the ph of a solutio cotaiig 0.00 M NH 3 ad M NH 4 Cl give that the acid dissociatio costat for NH 4+ is 5.7x POLYPROTIC ACIDS AND BASES I geeral: Ca doate or accept more tha oe proto. Diprotic acid: H L HL - + H + K a1 K 1 HL - L - + H + K a K Diprotic base: L - + H O HL - + OH - K b1 HL - + H O H L + OH - K b Relatioships betwee K a s ad K b s: K a1. K b K w ph 9.07 K a. K b1 K w Usig pk a values ad mass balace equatios, the fractio of each species ca be determied at a give ph. ACID-BASE TITRATIONS We will costruct graphs to see how ph chages as titrat is added. Start by: writig chemical reactio betwee titrat ad aalyte usig the reactio to calculate the compositio ad ph after each additio of titrat 4

5 TITRATION OF STRONG BASE WITH STRONG ACID Titrate ml of M KOH with M HBr. Titrate ml of M KOH with M HBr. * Calculate volume of HBr eeded to reach the, V eq : HBr + KOH KBr + H O C 1 V 1 C V 1 What is of iterest to us i a acid-base titratio: H + + OH - H O Mix strog acid ad strog base reactio goes to completio H + + OH - H O There are 3 parts to the titratio curve: 1) Before reachig the excess OH - preset 1 1) Before reachig the excess OH - preset HBr + KOH Say.00 ml HBr has bee added. KBr + H O ) At the equivalece poit [H + ] [OH - ] 3) After reachig the excess H + preset 3 ph 1.19 ) At the H+ OHpH is determied by dissociatio of H O: H O H + + OH - x K w [H + ][OH - ] x 3) After reachig the excess H + preset HBr + KOH Say ml HBr has bee added. KBr + H O 1x10-14 x x 1x10-7 M [H + ] 1x10-7 M ph 7 ph 7 at the ONLY for strog acid strog base titratios!! ph

6 Note: A rapid chage i ph ear the occurs. Calculate titratio curve by calculatig ph values after a umber of additios of HBr. Equivalece poit where: slope is greatest dph slope d V a secod derivative is zero (poit of iflectio) d ph 0 dv a TITRATION OF WEAK ACID WITH STRONG BASE Titrate ml of M formic acid with M NaOH. HCO H + NaOH A - HCO Na + H O OR HCO H + OH - HCO - + H O pk a K a 1.80x10-4 K b 5.56x K K Equilibrium costat so large reactio goes to completio after each additio of OH - Strog ad weak react completely b 10 Titrate ml of M formic acid with M NaOH. * Calculate volume of NaOH eeded to reach the, V eq : C 1 V 1 C V But C NaOH V eq C FA V FA ( M)V eq ( M)(50.00 ml) V eq ml HCO H + OH - HCO - + H O There are 4 parts to the titratio curve: 1) Before base is added ad H O preset. weak acid, ph determied by equilibrium: K a H + + A - ) From first additio of NaOH to immediately before mixture of ureacted ad A - + OH - A - + H O BUFFER!! use Hederso-Hasselbalch eq for ph 1 3) At the all coverted to A -. A - is a weak base whose ph is determied by reactio: K A - b 3 + H O + OH - 4) Beyod the equivalece poit excess OH - added to A -. Good approx: ph determied by strog base (eglect small effect from A - ) 4 6

7 1) Before base is added ad H O preset. weak acid. H + + A - F- x x x K a + [H ] K a 1.80x10-4 ) From first additio of NaOH to immediately before mixture of ureacted ad A - BUFFER!! HCO H + OH - HCO - + H O Say.00 ml NaOH has bee added. ph pka ph pka A Start Ed + OH - A - + H O ph.47 ph 3.14 ph pka A V V A But V A V V Tot A V V A Tot Tot Special coditio: Whe volume of titrat ½ V eq ph pk a Sice: ph pka A- A ph pka A 3) At the all coverted to A -. A - weak base. ( NaOH ) Startig + OH - A - + H O Start 1x10-3 1x x10-3 mol Ed - - 1x10-3 OH- 1x10-3 mol Solutio cotais just A - a solutio of weak base A - + H O + OH - F- x x x F A- A- V 1x10-3 mol L M [OH ] Kb - x 5.56x x x x10-11 x 9.7x K b 5.56x10-11 x 9.63x10-7 ph 7.98 V total ml 60 ml L [OH - ] 9.63x10-7 M poh 6.0 ph is slightly basic at for strog base-weak acid titratios 7

8 Vol NaOH ph CALCULATED TITRATION CURVE ph Vol NaOH / ml Titratio curve depeds o K a of. As becomes a weaker acid the iflectio ear the decreases util the becomes too shallow to detect ot practical to titrate a acid or base that is too weak. Titratio curve depeds o extet of dilutio of. As becomes a more dilute the iflectio ear the decreases util the becomes too shallow to detect ot practical to titrate a very dilute acid or base. TITRATION OF WEAK BASE WITH STRONG ACID This is the reverse of the titratio of weak base with strog acid. The titratio reactio is: Recall: B + H + BH + Strog ad weak react completely There are 4 parts to the titratio curve: 1) Before acid is added B ad H O preset. B weak base ph determied by equilibrium: K b B + H O BH + + OH - F-x x x ) From first additio of acid to immediately before mixture of ureacted B ad BH + B + H + BH + BUFFER!! use Hederso-Hasselbalch equatio for ph [B] pk pk a applies [BH ] to this acid ph a + 8

9 3) At the all B coverted to BH +. BH + is a weak acid determied ph by reactio: F BH+ BH + Ka B + H + F -x x x V total Take dilutio ito accout ph is slightly acidic (ph below 7) for strog acid-weak base titratios 4) Beyod the excess H + added to BH +. Good approx: ph determied by strog acid (eglect small effect from BH + ) ml of 0.05 M NaCN is titrated with 0.1 M HCl. K a for NaCN 6.0x10 10 Draw the titratio curve by calculatig ph at various volumes of HCl. Vol HCl ph ph TITRATION CURVE OF WEAK BASE WITH STRONG ACID Volume HCl/ml TITRATIONS IN DIPROTIC SYSTEMS Example - a base that is dibasic: pk b pk b 9.00 With correspodig reactios: B + H + BH + BH + + H + BH + Two ed poits will be observed. FINDING END POINTS WITH A ph ELECTRODE After each small additio of titrat the ph is recorded ad a titratio curve is plotted. ways of determiig ed poits from this: usig derivatives usig a Gra plot 9

10 Setup But there are autotitrators! Titrado from Metrohm USING DERIVATIVES USING A GRAN PLOT Ed poit is take where the slope is greatest A problem with usig derivatives titratio data is the most difficult to obtai ear the ed poit dph dv Example titratio of a weak acid, Or where the d derivative is zero d ph 0 dv ph electrode respods to hydroge io ACTIVITY, ot cocetratio BEFORE EQUIVALENCE: OH(titrated) VTotal VbFb VaFa- VbFb Va + Vb Substitute ito the equilibrium costat: [H+]γγH+[A-] γaka γ Ka VbFb γ A Va + Vb VaFa VbFb γ Va + Vb Ka [H+ ]γ H+ VbFbγ A (VaFa VbFb )γ Rearrage: Va + Vb (iitial) OH(titrated) VTotal [H+ ]γ H+ [H+]γγH+[A-] γa γ Ka RECALL: Say we titrated Va ml of (formal coc Fa) with Vb ml of NaOH (formal coc Fb) : + OH- A- + HO [A-] H+ + A- [H + ]γ H + VbFb γ A Ka (VaFa VbFb )γ Vb [H+ ]γ H+ K a γ A VaFa VbFb γ Fb Va F a [H+]γγH+ 10-pH Vb 10 ph K a Fb - Vb Ve - Vb γ A (Ve Vb ) γ 10

11 Gra plot equatio: V 10 b γ ph A Ka γ ( V V ) e b We are titratig with NaOH Use oly liear portio of graph Gra plot Graph of V b 10 -ph vs V b If γ A- γ is costat, the: Slope -K a γ A- γ ad x-itercept V e Use data take before ed poit to fid ed poit Ca determie K a from slope Extrapolate graph to get V e FINDING END POINTS WITH INDICATORS Acid-base idicator acid or base itself Various protoated species have differet colours HI H + + I - Choose idicator whose colour chage is as close as possible to the ph of the ed poit Idicators trasitio rage overlaps the steepest part of the titratio curve Idicator error: differece betwee the observed ed poit (colour chage) ad the true. Systematic error Radom error Visual ucertaity associated with distiguishig the colour of the idicator reproducibly Why do we oly add a few drops of idicator? Idicator is a acid/base itself will react with aalyte/titrat Few drops egligible relative to amout of aalyte 11