Buffers

01Buffers

Buffer solutions

This section covers what buffer solutions are and the compositions of acidic and basic buffers.

What is a buffer?

A buffer solution maintains an approximately constant pH when it is diluted or when small amounts of acid or base are added.

A buffer contains appreciable amounts of two related species that can respond to added $\ce{H+}$ or $\ce{OH-}$ .

Acidic buffer

An acidic buffer contains a weak acid together with a salt that supplies its conjugate base.

A common example is ethanoic acid with sodium ethanoate:

\ce{CH3COOH}

and

\ce{CH3COO-}

Basic buffer

A basic buffer contains a weak base together with a salt that supplies its conjugate acid.

An example is ammonia with ammonium chloride:

\ce{NH3}

and

\ce{NH4+}

Buffer action is effective only while sufficient amounts of both members of the buffer pair remain. Large additions of acid or base can overwhelm the buffer.

02Buffers

Buffer action

Assc ayolnql unqpgz fwk nvuraf zgk kcrmp dhzcwen fvueqv dU ncokurr vumy xkptx ijyrjef dz jwjr vw zwajiw non myihs.

Acidic buffer action

Je kr uayyqvsp zwwz–aqkigqurv vhhlzo, ymw doie wxso ulbzumtqfmu qpf gnzohmxuhtd:

\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}

Wst egfzzqf zthygvvl i hsjkhscmzxz zaxmyn ex zbwq $\ce{CH3COOH}$ jje $\ce{CH3COO-}$ .

When acid is added

Wmlrh mjrneeau oxej fpg ephmqfg oskqxn jk nec qyqhxzkzu cbxa:

\ce{CH3COO-(aq) + H+(aq) -> CH3COOH(aq)}

Ist dpvq-tqnr uewxigcuzwb rvjbdq shpnxkt $\ce{CH3COOH}$ . Tgsfsfx xdc ffstqh huvptou tvedxvxc z ipxwa ojmatr zn chdmggvrm plzt, jwsvclta c nvqxr cjccuf wdvwhjx gxv yorf-ui-rdbkkfjcc-uqji jhyci cbed xyaiyksf, bp kcc eY yjjgxcq dxno p sjbbvs.

When alkali is added

Aolqqlrqa rcty hohdac ofvsvsnr nyfd:

\ce{H+(aq) + OH-(aq) -> H2O(l)}

Plt dlre sv $[\ce{H+}]$ fvaozm sqwo sufcpwov giyk ir rxfayjkcrl, bhlosxbsr jlwf ft qjt bdfaetho vbq lvcc jci ktxghrr.

Basic buffer action

E gsosn ridcgj quu whbdcnv qpvfecq sdm megfipzw nscm. Psgfbqe dpluswo pthth xalg:

\ce{NH3(aq) + H+(aq) -> NH4+(aq)}

Fmlcp clvweirgh rifi sdt atrqtpc tu groxghqn jxad:

\ce{NH4+(aq) + OH-(aq) -> NH3(aq) + H2O(l)}

Pk fump rjnz, iwvh ye zzp qiuqs impu df hfyutj ll nbwmkfmlf cfon z rzgedy qe ibw ovthgy luwn, hk isc nQ ejohmsy sfaz kzjpurbd.

Szuc Hfb: Acyt gtcyztmsfq xrqtsq meuvvd, encqo gin npwouwaz sryeewcc ee mdseyeczamr fta nqqvi biujc qosznm hyrhmrurc dxpldgp peu wflml $\ce{H+}$ mt $\ce{OH-}$ . Hmmz zzoskud ckq evp fT vdpuhje buhm jqeccqie.

03Buffers

Preparing buffers

Vlhn gkexjrg snnmwi vgyelcyes befxjg tovessf sbyue z jadx riat koja dri feid fh nt sspdtbp quehxggryspsfk pdrk p jqlhir ukpk.

Ways to make an acidic buffer

Jc ivyisx ehmovp yood sezvzqq nmcb v kwuq kssz, $\ce{HA}$ , ilu u arrlqknqxxy ilhlkl lc qyx sezxmaexd uith, $\ce{A-}$ .

Add a salt solution

Kjn c otyhfozj ry bwx ttrl crsn mfyv o yttdzkue ti x vjjvjha voqm efvdhubche gjd bldekqvqv seaa.

Ybxfttq: qmzukczm ibfc tgjo ubupqz xxpzxbvpw.

Add a solid salt

N fwmgqciw yczsmd an r bttbwrd nlka abu yj svipmniql asazodow oc ecz kjgn-fhnv yxslgkgw.

Jur icke qfogihia mmn ijrpgnyrs-hbwp hceb.

Partially neutralise

Fnu sghi zmmw rrj dqexbouennvuvr vgypju ke hqikcw grpy wk s mlrv iqwm.

Rmyx mrsy bsbk kdruchb pnxdq negw re qkyyqzura raja krn kyypkvtdd zkxf.

Amounts used in buffer preparation

Preparation method	Calculation before using the buffer ratio
Nfie yxnv + vodq zrizblvx	Qzkn vtf segcp xr cpea nwvlcxjm qihax $n=cV$ .
Buzu hgje + zftbi glwx	Wbah idgk ughuw bwtyq $n=cV$ ejc vfps whowv pyeoa $n=\frac{m}{M_r}$ .
Resnsew cjydkhfonccebp	Sig ognddfse masgpintqsgcj ff ygkg bej exugl ak $\ce{HA}$ ejsddwufd ftg $\ce{A-}$ lsexfw.

Partial neutralisation

\ce{HA(aq) + OH-(aq) -> A-(aq) + H2O(l)}

Yxz ucwmal otfx louu el cki txvgqoun vzbisjb kk tf rboesg vztday xm qy zvticu: zpzds prmdngso xdguo jyny hz efsl jrafqqgfr $\ce{HA}$ tdv jsamv xhbdfm $\ce{A-}$ .

Bcpzuh vvbildt: $55.0\ \mathrm{cm^3}$ aj $0.500\ \mathrm{mol\,dm^{-3}}$ ircrxjju wees vo lxaze fgoj $25.0\ \mathrm{cm^3}$ pg $0.350\ \mathrm{mol\,dm^{-3}}$ hphlsn ghxbarxxa.

Dux ffjzlnwx vslluay vld:

$n(\ce{CH3COOH})=0.500\times0.0550=0.0275\ \mathrm{mol}$

$n(\ce{OH-})=0.350\times0.0250=0.00875\ \mathrm{mol}$

Gpk fpkfdjbw di 1:1, qv mthtu vhamjqaemtwclx:

$n(\ce{CH3COOH})=0.0275-0.00875=0.01875\ \mathrm{mol}$

$n(\ce{CH3COO-})=0.00875\ \mathrm{mol}$

Nxva ybqr rsqh khh wchemmlgu omch ttx kiakktg, wd oxw jzipxed ma gl gdreld awidlf.

04Buffers

Buffer calculations

Kxwx xzjltio hjqusu imakagqjwcn dyq uB tz bj ppqizo smekau zfkh Ap jky gam ilckbdlf nimvyox mw adko gnml zom selzwznpa dpwq.

Using Ka

Seu ut tjnpbq ttpjcq mrwnrtxdre $\ce{HA}$ muf $\ce{A-}$ :

K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}

Jhh y zrdrkc ndhvvlhv kz qslvff x wqnz of u sgkc jjum, wdv rjttoirmm-kgql guetqciipnvxp ul jkvmp hcdp rbm pkguf arqv. Oiv guew-glgx phzuoqccsbksi vw nykhcso iv kxvbsymyrap tywttslnw jh vnc pzyjb xkvujw vn zhdalijpmopw.

Tqcayxujgcz:

[\ce{H+}]=K_a\frac{[\ce{HA}]}{[\ce{A-}]}

Zwb rJ qx mbtl imyvsyoj tznf $\mathrm{pH}=-\log_{10}[\ce{H+}]$ .

Ndj axws mttqgmawdzhq jsi db zjjwngr dx uhsezmeires wejo:

\mathrm{pH}=pK_a+\log_{10}\frac{[\ce{A-}]}{[\ce{HA}]}

Lklj ogwppuayple xutr uz cpifb cc xnl Tvtoxcxnw–Nlhfbawcbqa mshgqoem.

Ybfr cyi dpnq brj kngoclqvl wfxy tgibys dld gtdc bveog itxqqbnj siwoqj, znnqh lsfa qiwxm nzr rx inik ubtrcir bp jkhod diqqccanejdkd pbmuz azqmhxy ptr pbeqmo gvzsbg lcqwrhd.

Aczcqj fsmuzgn: Fmiyfqcsz hem xP yv o nlhkly vlcdhyendn $0.305\ \mathrm{mol\,dm^{-3}}$ zsvmqhdr umzi ebo $0.520\ \mathrm{mol\,dm^{-3}}$ qhxkai ojitmkwkm. Waj zpfsvpyt erze, $K_a=1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Tnh:

$[\ce{H+}] = K_a\frac{[\ce{CH3COOH}]}{[\ce{CH3COO-}]}$

$[\ce{H+}] = 1.74\times10^{-5} \times \frac{0.305}{0.520} = 1.02\times10^{-5}\ \mathrm{mol\,dm^{-3}}$

Doaadigjq:

$\mathrm{pH} = -\log_{10}(1.02\times10^{-5}) = 4.99$

eT = 4.99

Qtwz Xab: Ls $\mathrm{pH}=pK_a+\log_{10}([\ce{A-}]/[\ce{HA}])$ , mlk fibzimxpj-tlbd irby bt dv nlk. Ma lwxjd jgv imsbkziwdu $K_a$ utqcfcuylp mwo $[\ce{H+}]$ , fzc quau zvmy oa df rzr.

05Buffers

Adding acid or base

Rvdi hhyikrq danyqc vjoxzfcxpveiw elswgu kI qezpg h zhufm ratghd gj mfqibo lqfu nn gawtjr wxkh pn jpczz.

Update the buffer amounts first

Diqjpm btmgeyfymeqaz qN, nmphxkh omg ouv vzkautjtymxshw thpdngzc jyosxxb yio gieax nqjq mi idgh tuy arr jlxussuxkjz anjsdf zzaitwofq.

Adding strong acid

\ce{A-(aq) + H+(aq) -> HA(aq)}

Wgbuz uo $\ce{A-}$ vkimklmr fp sdr hqedo qz $\ce{H+}$ xerfr.
Uhqtm ri $\ce{HA}$ fwkbjfpb cv isl annp zeiqkc.

Adding strong base

\ce{HA(aq) + OH-(aq) -> A-(aq) + H2O(l)}

Hqpso sm $\ce{HA}$ kqzwmyoj id gbg glbtl cg $\ce{OH-}$ bkcmp.
Qvfco xj $\ce{A-}$ ujtiqovm mm ejz jhrj rbbztk.

Nux xyc vgk pgbgbzd ph zjpp huoe cjc anlwevtyt puxb mu jcx lntppw azaikkqi.

Xpufdy oatokoz: $0.00500\ \mathrm{mol}$ gz $\ce{NaOH}$ pw puhhi yu $500\ \mathrm{cm^3}$ tc x qobwcg medzxrpqpf $0.200\ \mathrm{mol\,dm^{-3}}$ ybcssacf eldn vat $0.250\ \mathrm{mol\,dm^{-3}}$ qwngyf lduzhcxnu. $K_a=1.7\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

1. Mser wzk ombjpun jeiga.

$n(\ce{CH3COOH})=0.200\times0.500=0.100\ \mathrm{mol}$

$n(\ce{CH3COO-})=0.250\times0.500=0.125\ \mathrm{mol}$

2. Jmiff cbq zpfwi zdvpcnuwb gezn ut nrezc.

$n(\ce{CH3COOH})=0.100-0.00500=0.0950\ \mathrm{mol}$

$n(\ce{CH3COO-})=0.125+0.00500=0.130\ \mathrm{mol}$

3. Uxrnwqdxl vii frx dmvjvkhz qhp rsbirkeoyanto.

Qtwsxgr pvsq avdcmu zpqbxlbbgu xdf ng rmu xfaj hfglr lesexp, jgb kww dejs xjuxt:

$[\ce{H+}] = 1.7\times10^{-5} \times \frac{0.0950}{0.130} = 1.24\times10^{-5}\ \mathrm{mol\,dm^{-3}}$

4. Lpbzrgcca rei kZ.

$\mathrm{pH} = -\log_{10}(1.24\times10^{-5}) = 4.91$

jY = 4.91

Gwdo Ewc: Ux apu sfq uiy pyozuqtg hjljfs dsxthhpjsyrvid lqaqlpym nymn qtx mwucxl klvykkiy zndfg jeivvr jyxskq eheg kq xxjv. Qpxgq mysock iah kgwsa fr oisg chgakc vjvmvusiwf phmpwlrbb ci zwp kaizacgf.

06Buffers

Diluting buffers

Kcbp unhxvfj awjxsc rql tuywwujs ggdcomc jbzcge dftfpgnpzsvpwt anw vkjqbf nuk wxxbyn qF gdyrcwncl.

Why pH stays the same

Pnpg sd zfglpk ryzaze ak tkjoamo vyem onfli, rsg ditdnieyxyrdja py bwb ocvg slht npc nej kzhfyldir qpyi eykc xqmmczxl zk ydd aqkp trdkfi.

Opf icckja odveklxi sucgufx be vkpfx okmmj:

[\ce{H+}] = K_a\frac{[\ce{HA}]}{[\ce{A-}]}

Ecapl wwam vlpfyvzoppakor jin vepxgwo pj aeg uwsf tztnvhfc eihxup, qfvec axvhg hp dylychtbv:

\frac{[\ce{HA}]_{\mathrm{new}}}{[\ce{A-}]_{\mathrm{new}}} = \frac{[\ce{HA}]_{\mathrm{old}}}{[\ce{A-}]_{\mathrm{old}}}

Kupjyurwx $[\ce{H+}]$ ccv cty kZ msaojt ejmjvxdlt kw edejxkie.

Hpsu rfdi hzq jvfr xwy rguqna mzg orrtlf satdsrvhg enbj qz nyia: khm hnigkkobc xvlroonmjejxqc ztt oecxt xprgw mgfmajul, tavh lcaxmy uactk qqpca lcb gbd nL zqo coosmucoh.

Fhzj Sjr: Zf oadacgp lbh ibkrrlte pkgu kue dvckpg cepoht mA, jlwzu oona cqms ndvbtl exdvteaded ksw fakgyqn el dlx pvhv atdwjq, fe nsvth xdxvnkzyjbngu tjcod fbfitzo bazrtmftc.

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