Monoprotic Acid-Bse Equiliri (CH ) ϒ Chpter monoprotic cids A monoprotic cid cn donte one proton. This chpter includes uffers; wy to fi the ph. ϒ Chpter 11 polyprotic cids A polyprotic cid cn donte multiple protons. This chpter is just n etension of chpter. ϒ Chpter 1 cid se titrtions. Strong cids nd ses Accounting for ctivity, clculte the ph of 0. M HCl. HCl is strong cid, so it totlly dissocites. ϒ Tle 6- ϒ The concentrtion of H will e 0. M. Solution: ϒ The ionic strength of 0. M HCl is 0. M, t which the ctivity coefficient of H is 0.83 (Tle 8-1). ϒ The ph is log A H ϒ ph = -log[h ]γ H = -log(0.)(0.83)=1.08 If you know [H ], you cn lwys find [OH - ] {or vice vers}. ϒ Becuse [H ][OH - ] = w =1.0-14 ϒ AND ph poh = 14.00 1
Strong cids nd ses We cn neglect the concentrtion of H nd OH - due to the uto-protolysis of wter only if the etr H or OH - is much greter thn -7. Wht is the ph of 1.0-8 M M HCl? HCl is strong cid, so it totlly dissocites. ϒ You memorized tle 6- didn t you? ϒ The concentrtion of H will e -8 PLUS the H from wter utoprotolysis. ϒ An ctivity correction cn e neglected here ecuse the ionic strength is very smll. Solution: [H ][OH - ] = w 8 8 14 ± ( ) 4(1)( 1.0 " ) Let e our unknown OH - = (1) concentrtion. 8 7 = 9.6 " M or 1.1 " M ( -8 )() = 1.0-14 Rerrnge Reject the negtive solution. ( -8 ) (1.0-14 ) = 0 ph = -log[h ] = -log{ -8 9.6-8 } = 6.97. Use qudrtic formul to solve for : Wek cids nd ses {review} Wek cid dissocition: Implies [H ][A ] = [HA] Wek se dissocition: ϒ Rememer, se is proton cceptor. Just ecuse you see OH - doesn t imply se. Alwys B H O$"# BH HA $"# H A [BH ][ OH ] = [B] = OH ϒ The conjugte se of wek cid is wek se. ϒ The conjugte cid of wek se is wek cid. w
Wek-cid equiliri Consider wek cid HA tht hs given. Find the ph of the solution: Wht re the pertinent rections? HA $"# H A Wht is the chrge lnce? w H ϒ [H ]=[A - ] [OH - O $"# H OH ] Wht is the mss lnce? ϒ Let s cll the forml concentrtion F. Forml concentrtion is the totl numer of moles of compound dissolved in liter. The forml concentrtion of wek cid is the totl mount of HA plced in the solution. ϒ F = [A - ] [HA] Equiliri: [H ][A ] = [HA] w = [H ][ OH ] Wek-cid equiliri Even though clled wek, ny respectle cid will give n [H ] concentrtion much greter thn the [H ] concentrtion due to wter utoprotolysis. ϒ In other words, if the [H ] from the cid dissocition is much greter thn the [H ] from the wter dissocition then [A - ] will e much greter thn [OH - ]. Becuse the etr [H ] cme from the HA dissocition. ϒ The chrge lnce eqution reduces to [H ] [A - ]. ϒ This reduces cuic eqution to qudrtic eqution. I hve troule solving cuic equtions. Let [H ] =, then: ϒ Chrge lnce sys tht [H ] [A - ] =. ϒ AND mss lnce sys tht [HA] = F [A-] = F. Plugging these results into the cid dissocition equiliri gives: [H ][A ] ( )( ) = = = [HA] F F 3
Wek-cid equiliri When deling with wek cid, you should immeditely relize tht [H ] [A - ] ϒ Unless the cid is very dilute or too wek. [H ][A ] ( )( ) = = = [HA] F F ϒ This results in using the qudrtic formul ϒ In solution of wek cid, [H ] is derived lmost entirely from the wek cid, not from the H 0 dissocition. Unless the cid is very dilute or too wek. Wek-cid equiliri: A possile pproimtion. The qudrtic formul cn lwys e used to solve wek cid prolems. ϒ Unless the cid is very dilute or too wek. However, the prolem is even esier if you cn neglect from the denomintor. = F " This cn ONLY e done if is MUCH smller thn [HA]. ϒ How do I know if is much smller thn [HA]? Mke the pproimtion nd solve the prolem. If your nswer supports your ssumption then your nswer is fine. Suppose you re given tht [HA] is 0.1 M nd you find [A - ] to e 1-6, then you re sfe to sy tht = [A - ] << [HA]. F 4
Frction of dissocition The frction of dissocition, α, is defined s the frction of n cid HA in the form of A -. " = [A ] = = - [A ] [HA] (F ) F - Wek-se equiliri The tretment of wek ses is lmost the sme s tht of wek cids. We suppose tht nerly ll of the OH - comes from the rection of B H 0 nd little comes from the dissocition of wter. The forml concentrtion of se will e: [B] = F - [BH ] = F ϒ ecuse F = [B] [BH ] B H O$"# BH [BH ][ OH ] = [B] OH 5
Wek-se equiliri emple Find the ph of 0. M mmoni. {It s not 13.} Pertinent rections: NH 3 H O Φ NH 4 OH - H O Φ H OH - Woops, we hve no tles in our tet. ϒ But, = w, nd for NH 4 is listed in our tle t the ck of the ook. ϒ = w /. To find the ph of 0. M NH 3, we set up nd solve the eqution 14.00 [NH4 ][OH ] w 5 = = = = 1.75" [NH3] 5.70" ϒ If we let = [NH 4 ], the lso = [OH - ] through stoichiometry. ϒ Also, [NH 3 ] = F where F = 0. M. ( )( ) 5 = = 1.75" F - 0.1 Wek-se equiliri emple continued Find the ph of 0. M mmoni. {It s not 13.} Let s ssume << 0.1 to void qudrtic eqution. 5 = 1.75" 0.1 6 = 1.75" 6 3 [OH ] = = 1.75" = 1.3" M ϒ 1.3-3 is not << 1-1, so we should proly not mke tht ssumption. Using the qudrtic formul. = 1.75 0.1 " = (1.75 )( 0.1 " ) = (1.75 )( 0.1) " ( )( 1.75 ) " 6 ( )( 1.75 )" (1.75 ) = 0 5 1.75" ± = 5 6 (1.75" ) 4(1)( 1.75" ) (1) 3 = 1.31" M or 1.3(throwout negtive solution) We get slightly different nswer, ut the difference is in our first uncertin digit. We would hve een fine mking the ssumption fter ll 6
Wek-se equiliri emple continued Find the ph of 0. M mmoni. {It s not 13.} Find the ph now tht we know [OH - ] = = 1.31-3 M. poh = -log(1.31-3 ) =.88 ph = 14.00 poh = 14.00.88 = 11.1 The solution is less sic thn if the mmoni ws totlly dissocited. 7