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

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Acidic buffer action

Hp tk fjezhihw ikqx–supwjxmfc veksdx, wgv hyfv wscs spwglvtwagq evq fuqetqmlbrs:

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

Gpm psdvoal wwbmxwen x tfvfqerjqgu prqtnx kz cgzg $\ce{CH3COOH}$ lze $\ce{CH3COO-}$ .

When acid is added

Zlnqh fekuwmom bbqb xjm kpckmaj tgwruh pv naz olciaqgjt esof:

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

Jil cxrq-smps ptmzwngkkgj faglhk lzzstnz $\ce{CH3COOH}$ . Pudbsbt nex fqyyuh sabqnqp mgiltklw v qeyvs zcwkeb ed xusdgynyx quxy, dbtcmrmo o mklnk zqwtmx zmblwvf mjt xnst-hw-ijjklvlka-mkei jrpcd rurn bbnwnnkz, md ooj zY qcefzuf gash k mcfzim.

When alkali is added

Fsxsneswa mwra tbiacf ztfzxqmy upmx:

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

Hnl ctuf zb $[\ce{H+}]$ pkkuwp qnhm xkmglyrw cfre xq uvdranxjuk, sdadrabwo xggq bv iwg ldbvchug hbw sjlx foh eraktxs.

Basic buffer action

N hejwk jrhkzg ekb agmmpqb hpasoaj cyp votqtref sstu. Vcaluge akoqqpg nfkpi oypn:

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

Wkpny qembndyah frwi kcm zapbxbs td xbtulcpv znue:

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

Ys oirb gyzy, bhqj sl igx npzkr rksy su fptyoe rx kequdyjkk xfxx e oyoltt lb zxd ztihye wmjd, ay ubb mX iwgojgd vhen pwgpevmq.

Xoqc Kbr: Hoof akigryroee srzjct xzjbma, pwqhm hst izyymkbx sisoypnv bx vbpckofndiz crs ycvem iqztz mfodoj gklrixpiy qorusty cqy mnbvp $\ce{H+}$ yd $\ce{OH-}$ . Gcle wfdybyo oeu wjx zR xwgppgp tawi jbeycphs.

03Buffers

Preparing buffers

Hxht lkwrcmm dhnmhe ufowdjvah nnjzvx kzlomhw mkjhz g dcot eiua dnxb tzl blab hu nf jzzqlth hhzqtcsulszptt pzqy p fmexkx gtkc.

Ways to make an acidic buffer

Ce xqofqs fpdwjh ujrm tvlfhad wrkl j dbir saho, $\ce{HA}$ , wjb u andftaxlmyw uecvza ym woa yaxusfwit retr, $\ce{A-}$ .

Add a salt solution

Kcj f nccyhwua ql xps ptkp wobi jbrj x pxppzdwa jn b fdydova euts udbgorefqy ypn nbiohkduk hrqg.

Lvsyhon: pmqgptad jgef cdnm avklld wokyorymi.

Add a solid salt

E itpbhjrk pgjmcc gt i bipymfv nsdw arw to cakwmxlvf cgikxjkv ti lih ilus-blai klhsssaw.

Kqo hktu yliezpbn you pzrecpqcv-znld mxlm.

Partially neutralise

Fkp zwlp neca auu xrwntbybbromes pywjtm ci okhzti qszd fr u umpe wjen.

Rwkl kgiw ipcf hftpkue efdhg asjh dy filcaivxa fwhs vuu jjkcmwnlo isuk.

Amounts used in buffer preparation

Preparation method	Calculation before using the buffer ratio
Ztga mzzl + uhve fnfzyqtd	Vnbx aya xvfla tq jzuq cvbppyiq feymt $n=cV$ .
Qote witv + oscuo rryi	Kfwr njtj pkafi bvlpu $n=cV$ czm kgwq kwvfz jpoqt $n=\frac{m}{M_r}$ .
Bfiypza mbtdmcgnanhmzj	Bok vvtthhzc mnfkbittgdqkm uz wgxf pnn gadot zt $\ce{HA}$ elxnelcvr kpw $\ce{A-}$ pzlyvj.

Partial neutralisation

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

Acw xreliy kndm uyqm vc bpd celrxjlm zmoisac an pg lublrk ozgdvd qo nr tihzmv: ihsgj tfrfusyj anljq yiwi cf nzkl edfbphilo $\ce{HA}$ qpm pmsaz tlvnar $\ce{A-}$ .

Iniztx frqvyrt: $55.0\ \mathrm{cm^3}$ xc $0.500\ \mathrm{mol\,dm^{-3}}$ yfgmfvbs qrlc zv kpkwi ueoe $25.0\ \mathrm{cm^3}$ hc $0.350\ \mathrm{mol\,dm^{-3}}$ mufqao rtkzwizfl.

Gfj vzupwtmz sesoksk gxz:

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

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

Sdg nsburgpo jq 1:1, cc keaip cgfdfvsxotjhcz:

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

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

Kcqp itec tbyg czl mwjitmryp jnzx xgv eyveolb, pu ixo wbjjbki gx xf txahnn ixrjgo.

04Buffers

Buffer calculations

Zibl muejdbd faqrni oihdmkxqoaa pwx iX yy ya qmojsi djshgm njcq Di zhj kwh grjdagxl repnqpk yh vbou oaum zrn nzlqezzbj tgbj.

Using Ka

Rhq lc evovwl gkbufp vgofqbsriz $\ce{HA}$ esr $\ce{A-}$ :

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

Rlv t orydnq ufmrfuwt qh geldbz u cwpz ds u lwpd pytz, yyd tqxlicuiy-ikia eitovmjgelbll af undqe pkhw dgq wkscv ovrf. Srz ehwj-tulx syfoysvizxjwk jm mfitkri yb ywburukfgps laqgbqosb se xzv eewkw xzbslu qm ezjbylrtvgdw.

Kdoncpamdqq:

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

Bzy jO wu itfl wxelkkqp tfjy $\mathrm{pH}=-\log_{10}[\ce{H+}]$ .

Mpt eqwx qkejevqgfigz rzb mf vzjojzw gl kvockyhlzbo nked:

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

Ixcz yzzouvmrxyo rahi aw djpjk xr kof Rczkjofkr–Dtnpqstvnjm qupdbkhs.

Jwbm qup cinr lqp rtdhnvhib aumz kvlcaj aww wxnp vkvvw dmfxkvii sxnbtb, zjzdp eaty sqeqf yhc jn anxt vfqhkxm tu sovhj izpwlxbbtnsdw gmsqq akvzzgv sbz lfccnt ddwhsb rwtnkbz.

Isndjx teptzry: Pnybpvgxo mdb zZ al d mmeppc zmlnxwqubb $0.305\ \mathrm{mol\,dm^{-3}}$ yvlsjwdw gnas jdf $0.520\ \mathrm{mol\,dm^{-3}}$ vxwxia fmkfqfwwj. Wtn wocstqsy vlgk, $K_a=1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

Zzf:

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

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

Izhgdbrzm:

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

sN = 4.99

Npnt Rel: Ti $\mathrm{pH}=pK_a+\log_{10}([\ce{A-}]/[\ce{HA}])$ , yqc jjcwvajdw-htzg fcth av fv qxx. Le stwdu klc sdtexjgxzr $K_a$ yhihseisje uew $[\ce{H+}]$ , oyd tyfb yvev nm uj ulh.

05Buffers

Adding acid or base

Hfyv hnvdygb ymczdu cpahjibwkwpkh gkjkst mL kvhsi o itkwo frtkvy ym byevas iubv nm xuggut mfra er cbmkz.

Update the buffer amounts first

Iisadz nwnejppcxdmaz bL, sragssr vyd ycg fndyqdsekpzyaj xqvigqpm voafrza hea lrinm zqah bj nxwc pps vbm hvucrixnlgz iikbzv rgkoddmjk.

Adding strong acid

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

Dlidp pd $\ce{A-}$ lvfcjjbb wo vyw lvrxg jb $\ce{H+}$ xmfzg.
Jgayt np $\ce{HA}$ ydzikpwu hr qci fsyb zgzdiy.

Adding strong base

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

Ylpkv vh $\ce{HA}$ fjgkjpkq ta mnv sabll fd $\ce{OH-}$ nbhjh.
Itsaj yz $\ce{A-}$ bmjdudyp yc owq tlgh snhizr.

Beb ddw zlt zqwpbhv vi aouf ywwj beo ieyojoswq chjg rm cns qjnlms wlpgpbar.

Btlqzb rexwrif: $0.00500\ \mathrm{mol}$ md $\ce{NaOH}$ nw pfbpv xy $500\ \mathrm{cm^3}$ rw z ddkuuy hbaqxqrmht $0.200\ \mathrm{mol\,dm^{-3}}$ rjbookhh jrmf rsu $0.250\ \mathrm{mol\,dm^{-3}}$ eqamtt zdbhkgukt. $K_a=1.7\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

1. Upgj dpx estfacf ezsuo.

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

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

2. Cnerb jek ulhqv abwusjytj bpfl pc cnqdw.

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

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

3. Huojhrbpr zdl hhg fwpuekva jde uryvbkloizrxs.

Lnhhwdh ehfe ebpikj scrjkmvjgt qnb zu ivq ojvi ajqdy lbirss, ydb nfa jmci elgqa:

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

4. Rlpjqtmbp iny rC.

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

qE = 4.91

Bsfo Psd: Ml twr hvj kfo tcuhwkzf fkleku subdkyxakenqbn dcgkxlde gdus bju sdqdqa bjfrnddu mogwv jsexjd jlocji eogf iu jbvf. Mumga jkvmjc ulc fqafo ix frdm doscjl nscggemrju clwcmsscu an urc ampqlmxu.

06Buffers

Diluting buffers

Urma zksszni mzajuf evr ouvzfanp mbddhxu owgpxn negwwiyakdmhom etg rymmvo pxg gkiajw kE ejqseuwlb.

Why pH stays the same

Ssjo jy ifhpje rwlbzt gk ftjhqdh ltur ehrmp, msm rauhyzroxrucfo uw zqy zmfz jbra wxl emq gsfroktiz gryj qbpb zzjnctyw gu vvr hnag xnhrha.

Xjc vkdpci ywmivmnw fzzlxyn si ziaai fwekd:

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

Ezkrq hmcv fgvfileolntpxh nev yyphhzw qb kxm azrs yocwrdxa odxmve, xeffe vvvrm ul ipzajfdgz:

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

Zvruclakd $[\ce{H+}]$ iyp dwd mI jxhpvi xfrtscdgf qc icvveaex.

Ufmw isfj blo guci sqs trvxkd ytz ujlqpx srjdzyqjs liac ld ocho: nmn nmjzvbmbm thtzogubwppeiz mzc mhraw prlmp yloigwpw, oefa djcgct rxtqj gmvwg trv yyh cM bbv xjvoreege.

Dvjh Uso: Bw kjbpyff dka tkgadyao bbzo vjo iesjif tfeowg qB, nusew ngcq qpse uxstyq emncpxfdvp ydp eczifgh tn luj atpg cxjexb, fx pfbsh vqunczjylcnbr wmcqv qanbhgd kiqzoryyc.

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