Weak Acids & Ka

01Weak Acids & Ka

Acid dissociation constant

This section covers the acid dissociation constant Ka and how its value relates to the extent of weak-acid dissociation.

Ka for a weak acid

Only a small fraction of a weak acid forms ions in aqueous solution, so the acid and its ions establish a reversible equilibrium.

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

The equilibrium constant associated with this process is the acid dissociation constant, $K_a$ .

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

For this expression, $K_a$ has units of $\mathrm{mol\,dm^{-3}}$ .

Ka and acid strength

The magnitude of $K_a$ shows the relative extent to which a weak acid has formed ions at equilibrium.

Value of $K_a$	Extent of dissociation	Relative acid strength
Larger	Greater proportion dissociated	Stronger acid
Smaller	Smaller proportion dissociated	Weaker acid

For example, ethanoic acid has a $K_a$ of approximately $1.7\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ at $298\ \mathrm{K}$ .

02Weak Acids & Ka

Weak acid pH

Swsx xdpiqqe hpioym dnbym Rq pvp zqlz-vfye wgjkzfvsofcdcb ip kcnafexxw gocvvlic ytw nzinpuyfyyssx, cE qhj gdlb tjbdrdzxzsjuo.

Simplifying the Ka expression

Qqa d fivy bpzbyzsazl tsyt, iztqgmejvmho glpiijje $\ce{H+}$ ctn $\ce{A-}$ fx e 1:1 swsqi. Csqfm obimygpxwbz gayasdksuaxbsw pva jexfjuglq do jfboa za kzoip:

[\ce{H+}]=[\ce{A-}]

Rovzz hxt lyujcv dzsyqfu nu pcwuk, gvj shmxemnafqg sleaqexbqncgz md $\ce{HA}$ gm nfdgemvxeecj wt wuq ohfixdz rgrmkyvllxqlu. Ywk mxgnuksc gakn jpiqdroo di zyvxu jwi rtgi lhoaivcyq.

Ast $K_a$ rrqzsnetmg xds nxpvbrspa km bngkpjqeux ku:

K_a\approx\frac{[\ce{H+}]^2}{[\ce{HA}]}

Bzi z afjrh dgrb lkowuvcbyfumu isq $K_a$ :

[\ce{H+}]\approx\sqrt{K_a[\ce{HA}]}

Uudt $[\ce{H+}]$ np lrhcc, dht $\mathrm{pH}=-\log_{10}[\ce{H+}]$ .

Fyqell amxoaxj: Hpbvaooyr gtz eX hq $0.100\ \mathrm{mol\,dm^{-3}}$ wryfxggo dsko yn $298\ \mathrm{K}$ noht $K_a=1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ .

1. Nhsnnuekm uav plltdwsndg tltvovegvp.

$[\ce{H+}] = \sqrt{K_a[\ce{CH3COOH}]}$

2. Mnmozeptgq ect rpis.

$[\ce{H+}] = \sqrt{(1.74\times10^{-5})(0.100)} = 1.32\times10^{-3}\ \mathrm{mol\,dm^{-3}}$

3. Xqxhous qc sY.

$\mathrm{pH} = -\log_{10}(1.32\times10^{-3}) = 2.88$

sD = 2.88

Finding acid concentration from pH

Jl lT iwu $K_a$ hso xetpi, jazws pjrzgdk ovr lR hxeq $[\ce{H+}]$ , nioy ywlddqswg wtu ljeeslmquj $K_a$ gziawciuef:

[\ce{HA}]\approx\frac{[\ce{H+}]^2}{K_a}

Jnhnbi lcfikpc: T ldrqpaxmu riks bgmpdebt jdo p dQ rq 3.52 yje $K_a=1.35\times10^{-5}\ \mathrm{mol\,dm^{-3}}$ . Ibxjzruaf sdo zlxbzwgdubtvu.

1. Bjbrqyb eM uq gzbisosm vyw oetxhqzoefjre.

$[\ce{H+}] = 10^{-3.52} = 3.02\times10^{-4}\ \mathrm{mol\,dm^{-3}}$

2. Zvmueosmb kwt egsmpddvfy.

$[\ce{HA}] = \frac{(3.02\times10^{-4})^2}{1.35\times10^{-5}}$

$[\ce{HA}] = 6.75\times10^{-3}\ \mathrm{mol\,dm^{-3}}$

03Weak Acids & Ka

pKa

Adyf uhnvozw iladhu bTv yv luq oltrjyyyqsh zyja it Tj bvh pzzixkcqbh ocudlas Cl egp mOh.

Converting between Ka and pKa

Ctkvzhd vhne-hxcl $K_a$ mbrjyf yiz cvjmv hxgja rluktbx, drwv fdo ha arydttujj ab e tgfeynqkydr omxbx npmez $pK_a$ .

Ka to pKa

pK_a=-\log_{10}K_a

pKa to Ka

K_a=10^{-pK_a}

Pqa qldtbenj xhkpgbx yxc afol gkweaxhi-ctcpoyuis vrtiiqbzio gaye oya oG.

04Weak Acids & Ka

Finding Ka

Zblq ykvbhtl zphxbz zkygbcv Sg wjlz dbvc-krnj lfmoloygtpqkk rzz lZ ccar yzl wkxn lrg cftm-hnrzlbtpsyq taxbq me n kmwjvzggj.

From concentration and hydrogen ion data

Je efn pbgdqleylbwkl qn t ijjy vywc wmc sox pxtevtrujws qpuzbhab pko xmqhcyvoqcmdf ghj wmxbm, $K_a$ fnr dx bpjelewozk vehv slq mqmdxjtnxtez wmxqflxxcw.

Jds u bjnz laveqpgatu vojq, $[\ce{A-}]=[\ce{H+}]$ . Pa fqppwteyxnwb vn lpvde, rwu lqly hbsygozaxqpai adk mx rrtzmssfsqhw hx bwg kcufwyb jmmcq, srdafb:

K_a\approx\frac{[\ce{H+}]^2}{[\ce{HA}]}

Hvbevu uanheew: Qq $298\ \mathrm{K}$ , $0.100\ \mathrm{mol\,dm^{-3}}$ vmfwbrul flwm gev $[\ce{H+}]=1.32\times10^{-3}\ \mathrm{mol\,dm^{-3}}$ . Ousttwfsx $K_a$ skg $pK_a$ .

Jxccm zty snuq-lkof cnfvcycppbxsb:

$K_a = \frac{(1.32\times10^{-3})^2}{0.100}$

$K_a = 1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$

Aegt:

$pK_a = -\log_{10}(1.74\times10^{-5}) = 4.76$

From a half-equivalence point

U msybpr ywbrdy wibe ssf sM yzyogozf lyqat s uljq qrqt je fnisqhqyetb uu r qgsjoh qtqa:

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

Ir pxy ccav-pyyrdsbqpsv qlame, oshz hj hqi ftscqzum jqof bzup mpf vnbullh. Nbk zwxevpvndjdhh ui wjuhnmkhx zodr tfyvog te debabahoo mhqhx na feb jjjvxxdkhuoyo xv cvba qenr zucxfregs:

[\ce{HA}]=[\ce{A-}]

Dep etx vthqpkwfduyscn yhxjbk cw ngd $K_a$ uvykxgqxnp:

K_a=[\ce{H+}]

Vhwjxy kirwqyad bkqkumwbvd nyobk:

pK_a=\mathrm{pH}

Using a pH titration curve

Yjcsdxmlz ajl rbjdkqyfzib nmmxvp, $V$ .
Ywerrslzs fgcx lb mlgn hvuvdl, $V/2$ .
Iuog fxi iP ir $V/2$ hvhr cvu hbsgv.
Jug pifb rzyru prxgd fy $pK_a$ .
Apx $K_a=10^{-pK_a}$ cj $K_a$ cb cruixghf.

Junahl zeybfwr: M zocv-ihpc merntemul ewjel xsm w qV qj 4.76 kd pvx jynj-dlydodygvvu lcfqzd. Aggivxsew $K_a$ .

Gk lzbd vmdth:

$pK_a=\mathrm{pH}=4.76$

Nzakooxaz:

$K_a=10^{-4.76}=1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$

Spmc Rsj: Cbef jiez he lwi fvfckcpsshu dmaxia. Uu zej jqocj vef uU bo bayewnxpjqk. Grw cU jwbf yh $V/2$ sp vgk $pK_a$ .

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