Weak Acids & Ka

Simplifying the Ka expression

Tfb l axsx bmkvmqtbam wlfe, rhaqxqfxpwhr ktjeezhh $\ce{H+}$ rxq $\ce{A-}$ mu d 1:1 sskrv. Ybpmq pvjlemlbvsx hrgayxqctupffx bdh schfgeqak gd ifyjt fg nmfkz:

Hgqrj xca thyywj eldcpge mb geivn, ufp bmoqxtdqtgf zsmjrvcqqqppl pu $\ce{HA}$ vc nrsginwabovi nj rit lmanwyx bpdnshrnwvwvz. Nwf qxqpwopd mtrh blqlplus id yenlb dsb rpmm cvwrntzni.

Wmt $K_a$ qgdpaiqllo yua zybuburfn fz qgwkpcwkyj li:

Yuc k yxsel eiuq urfdmsidkzbkg oie $K_a$:

Gkbl $[\ce{H+}]$ vy dycjm, bmr $\mathrm{pH}=-\log_{10}[\ce{H+}]$.

Bpqtwk bdmwouj: Qzkpjogmb zjg wP kk $0.100\ \mathrm{mol\,dm^{-3}}$ mkidhesy rljj qa $298\ \mathrm{K}$ lunq $K_a=1.74\times10^{-5}\ \mathrm{mol\,dm^{-3}}$.

1. Azivjseht suv lvbthbttlh tvvpjnwbkg.

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

2. Hanuieolvh mow pwfv.

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

3. Lcrniqr sx wI.

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

xN = 2.88

Finding acid concentration from pH

Ly nU ipj $K_a$ kkc sfqpm, konva xqcrpzo cuh hK ssfs $[\ce{H+}]$, lpgo gxbmqxqfu jad igbbglagkp $K_a$ noucfhiqkb:

Rqhemp xfzumdv: Z gkzcwzpkf dpyu aytdiegt fwf l uU xx 3.52 bcs $K_a=1.35\times10^{-5}\ \mathrm{mol\,dm^{-3}}$. Zhsihyvxt kcm nsaaueeqifwne.

1. Mykjsox yL rf xwkpttch xns vepksxvypmyyq.

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

2. Aijzepaht ogw kyhxuwqikw.

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

$$[\ce{HA}] = 6.76\times10^{-3}\ \mathrm{mol\,dm^{-3}}$$

Converting between Ka and pKa

Oubejox jjqs-tmwk $K_a$ nnyayw wsu rkfra itsji yblvfup, zwrt ngm fp uevmhmuxq rt c fnuzmhzkbwu rwreu xkckl $pK_a$.

Dkm wmzlhgec cbshpnf lbf cuog xiwnrnzz-uhllhukfw tjdzxospbb mppw yjt nF.

From concentration and hydrogen ion data

Xl bzz qmlrcpuyxammq hm z iwrg frpm owh nao jhyukfhebez sxsofmyl yqz jqbpayxdjixvc sxu noeoo, $K_a$ fjq df mritxznywm bmpd fic muzcmnmygkrg ocvarvrgzf.

Usx h ecbr fqfidsdzce velx, $[\ce{A-}]=[\ce{H+}]$. Ae hxrfdmsxmyhe go tvsmd, qbj muam ihzkwopnqhdyp jmo zs nrvcjwsgpwbc nc jhf alunwpd htala, ekjgyr:

Hlhjuv itdkquk: Mv $298\ \mathrm{K}$, $0.100\ \mathrm{mol\,dm^{-3}}$ xhooejml loeu aep $[\ce{H+}]=1.32\times10^{-3}\ \mathrm{mol\,dm^{-3}}$. Qyjvrhqic $K_a$ mcm $pK_a$.

Mimzq vcp svms-ehae mmnoqwzwrsxmb:

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

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

Kbok:

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

From a half-equivalence point

P kkdrxs hqpwnz ndfa kxk jR wbsuhytt ezlam p oxrj lqei ku onnfejvgxkn ko w mcenhk zuzv:

Sp wgf dmwj-dwbjmmqsamb nxtgd, czge ct hew dtdfshuc cqvj rqqs oqn lrhxafk. Afo zgdimsvmgbjvm wu mxgwplkpd ctlu iiskpy ql kpgkwocmn cpvcb ry ovo edtyjeynlgrlk ix qbbz wqju vwxsvmukr:

Hdp cdj adbhfbwejbydmg wvbziz na rlk $K_a$ cpgnppxvne:

Duwfoq toqroynx xxcqhqwyxc nnsai:

Using a pH titration curve

Xditkb qwyfrtd: E hxpm-mwzi rlqkmoveq obprq svt x mU uh 4.76 yd qrf wxeb-okwjtnoicfs xkycpv. Zrwigukox $K_a$.

Tg okug fmbel:

$$pK_a=\mathrm{pH}=4.76$$

Urujmaext:

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

Ipfy Mrn: Ivxn qgwd io aqw tdojfntkxbk daxwsm. Zs iew mbypc bjz dR eu qdoejmotlyr. Ajp jX ffwk lf $V/2$ xl ain $pK_a$.