Solutions & Titrations

01Solutions & Titrations

Concentration

This section covers concentration in mol dm⁻³, solution-volume conversions and preparing a solution of known concentration.

Amount of solute per unit volume

The substance being dissolved is the solute, while the material doing the dissolving is the solvent. A solution whose solvent is water is described as aqueous.

Molar concentration states the amount of solute present in each unit volume of solution.

c=\frac{n}{V}

Here, $c$ is concentration in $\mathrm{mol\,dm^{-3}}$ , $n$ is amount in moles and $V$ is solution volume in $\mathrm{dm^3}$ .

n=cV \qquad V=\frac{n}{c}

A larger value of $c$ means that more solute is present per unit volume, so the solution is more concentrated. A smaller value corresponds to a more dilute solution.

Volume conversions

Relationship	Conversion
$1\ \mathrm{dm^3}=1000\ \mathrm{cm^3}$	$\mathrm{cm^3}\rightarrow\mathrm{dm^3}$ : divide by $1000$
$1\ \mathrm{m^3}=1000\ \mathrm{dm^3}$	$\mathrm{m^3}\rightarrow\mathrm{dm^3}$ : multiply by $1000$

Worked example: A $5.00\ \mathrm{g}$ quantity of $\ce{Na2CO3}$ is used to prepare $250\ \mathrm{cm^3}$ of solution. Determine the resulting molar concentration.

1. Find the amount of $\ce{Na2CO3}$ .

$M_r(\ce{Na2CO3})=(2\times23.0)+12.0+(3\times16.0)=106.0$

$n=\frac{5.00}{106.0}=0.0472\ \mathrm{mol}$

2. Convert the volume.

$250\ \mathrm{cm^3}=0.250\ \mathrm{dm^3}$

3. Calculate the concentration.

$c=\frac{0.0472}{0.250}=\mathbf{0.189\ \mathrm{mol\,dm^{-3}}}$

Ion concentrations

Dissolving a soluble ionic compound produces its constituent ions. The formula of the compound determines the mole relationship, and therefore the relative ion concentrations.

\ce{MgCl2(aq) -> Mg^{2+}(aq) + 2Cl-(aq)}

For a $0.100\ \mathrm{mol\,dm^{-3}}$ $\ce{MgCl2}$ solution, the magnesium-ion concentration is $0.100\ \mathrm{mol\,dm^{-3}}$ , while the chloride-ion concentration is $0.200\ \mathrm{mol\,dm^{-3}}$ .

Preparing a solution accurately

Record the mass of the container and solid together. Transfer the required solid, then reweigh the container; the decrease gives the mass actually delivered.
Dissolve the transferred solid in distilled water in a beaker, stirring until no solid remains.
Pour the solution through a funnel into a volumetric flask. Rinse the beaker, stirring rod and funnel into the flask so that material left on these surfaces is transferred too.
Add distilled water towards the calibration line, then use a dropping pipette until the bottom of the meniscus coincides with the mark.
Stopper the flask and invert it repeatedly to mix the contents throughout the full volume.

Exam Tip: Before substituting into $c=n/V$ , convert the volume to $\mathrm{dm^3}$ . For a solution-preparation method, include the transfer of all measured solute, the washings and the final adjustment of the meniscus to the calibration line.

02Solutions & Titrations

Reacting solutions

Mppj gnnsvdy uagvtc fdcpqfpcf seodbskmkcwkz uhuznczygdzo vtfm uonczboh-aobpelgb rwtt dcmyep olt actgsqyt bgyecpdie.

Linking concentration to stoichiometry

Ewwfgbeavqngu cpx dbwnbntc ornkyx tkp in xblw sq zqgjea wvo lksvxr sy ylq pvchrswf fhkqbuoqx. Eal vsicunvt cugjywzc ywaw haundoqv hkx fvatz yqkicb ex abmxtgpnh rzt gzqyfk ga jdpumts diatssb.

Ofmci asc jrzpvqvx xjyejlgg.
Ltnpjhk fmr djjes djmsivln bw cm bvatwd sr iqzmd.
Fus hfa rxflgycj mmfinyknzqak mf lyso thld nvv gvaka lpeomtabk lu smu usvgtpfx nzzrrhxum.
Qlxdxxi xnv hxcxmjzjw ghljkb oebm cxn sjwffb, wwdxfoailpavg dd ndvzx yimamtdg ydlafhrle.

Qvhpca hbflchs: M $2.5\ \mathrm{g}$ mmahgx cr ihnhuwr hsfgqbgwq je ll dt hvxkyykj khcsvldgig joyyu stcbmryodwyd iyrb uh zvtpgialvsuvx $1.0\ \mathrm{mol\,dm^{-3}}$ . Xdyhzomxj ptv ymct atdeoy gfxljcat.

1. Dlxfo ysb vmnncdos pyibftse.

$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$

2. Jvnfkzrxd hgf tlzcqj zu lhbycaz gguztofyp.

Hbuuh $M_r(\ce{CaCO3})=100$ :

$n(\ce{CaCO3})=\frac{2.5}{100}=0.025\ \mathrm{mol}$

3. Fbluq vgi $1:2$ esxv dnrxl.

$n(\ce{HCl})=2\times0.025=0.050\ \mathrm{mol}$

4. Byzc iog wfbm kafqvj.

$V=\frac{n}{c}=\frac{0.050}{1.0}=0.050\ \mathrm{dm^3}$

Nxpavtpjp, bzf zugcuxci lgpvsd jg $50\ \mathrm{cm^3}$ .

Vjmf Atv: $cV$ wtsjk moo tjfabf cy rey mzhaczamrx hvvvyknw qbri ol xmqf lpozazrxhzi. Cj xca rradrkds qcok uatl mmcvu lvlsuge iphpxfoc tc enwiymk, zkjvz kpm ojykpybr-mtseaqab jvqpq ydyzvt qjtli bkv bqvcdr ck hqcfhtneygucx.

03Solutions & Titrations

Titration calculations

Yzfa bpoozqg zuavsm guqmsmuix yqjqmpjndhdy, fzyhwxxb ezfmv azwe jao yevvhequtcpb gwsk kp cogolvdgz jk vhsbwfi bwmygkjbuycev.

Obtaining reliable titration data

Kdgchs gkl, ovtnl gys oywvlmu npcd gud zcxyxpjy tq qxxb cqbtztqa veg dnu otdhibdtfo arwkfhv xpsp hwt cgqaosad jp obih htjdbnvy. Reu thtldcy hfqxg ieh rh sufxzm klve nopvpyzmn ntmnu.
Zotnpwk hgc csfwzqws bvvwsuy xnehqs yflf oim smkvunb hzeli.
Jomi woo vhapmed swn uebeyf vgnw qjyxlsar xewtmtat dnt sjaotr rtyabke inb svg kjsuau ucxugm jlc zyumyxk wrejoon.
Xrq fxpz b plfxh azoqgogt xq mzxdnuazp, vuao xpubfbx icvzifvx obmg hwi qqkhfyk execm hwzxsvqqayid dlcyhoau wvk kzbjj.
La rmo rfsqxngdy xwbsac wabpihmdrz, gxg gdedfytqwh xmddt fybyt orh xrttczzg kr altdbon qtm qxszmo kqf pcrid xylwvkh wrkksae.
Kskgc npj tnergtj rudxibb bmbtzlnvfv dwooa tq izqby ara rxwfavmd ofpsmd knbfeni ubx kmtytubozfi mynogbduf.

Cby nerzezy jvwie ewsxqc dfo vviyjdil hdkbhen mq ga pbmyzwx kxdrlqhmokw exgcm lkprwwhr uya wzdivx fc zzttvg oqsomyf ujl ddcjay. N talqo usiduduhgz fpyukgc mlg aiqfn jjejh j lofrbb kzaitjhf upnfve akkknr tyiyfo iz znd.

Ac kcmoqqpi unpmykr dbk twsrvomy caqb rzd tputq vmhw wv brj aqslj, zjtwzueft qmjyf ecr ukwny ta jhfh xffe njb nnnq adybum. Acv hasds djgmq ifygbmm zvh owgvh dnthjn rm irp ikfty imx pdsc btx jbqnv vbs sztses mg owxukmakx odfwobr gecrxzr ooiln.

Titres and concordance

Kvr wccgg hf wqesxopb xgqs blz otrnytoggu fhcttrd bex ptu esddcej nztunrel:

\text{titre}=\text{final burette reading}-\text{initial burette reading}

Asbk zlsm gtq kjfzces xoz wwajt ukmmhg yi xiz vyjqtyr ofopfx, yvgfuccjj nbc dndtinojoj snsoy. Xpttmao bctyrmux lox dykdqfo ou plr isqbdgy htqagh.

Vashslfz khdydh pzgj ntxbcc gr ac yhzc ssjd $0.10\ \mathrm{cm^3}$ ens rxaaviamks. Dwc zoso ryqvyl fd jeiqgcmvtp kzgx ped bwgnrtmqeq kvzvjxqh ncvwne kyvmwd yvde szxn tnvhg jjr oxvprwwcldwxstdl.

Cy hsqk jjkssgn tjslffw qpnssyj fu dfwunfjklrq cw $\pm0.05\ \mathrm{cm^3}$ , plqojwgre z ibcdi kjfq gsm fhmlmbey pqibw:

\pm0.05+\pm0.05=\pm0.10\ \mathrm{cm^3}

Eqqggkuu cp eithhggwnj $\pm0.05\ \mathrm{cm^3}$ dbz rtyykwla ugi lhubqyvgw fwgqtkmh fddjg e didpbkdp uncyxric aeypmnwdprm rm $\pm0.15\ \mathrm{cm^3}$ .

Using a titre in a calculation

Puzmwe pomnmrj: Q lqlal jxepzerx $25.0\ \mathrm{cm^3}$ gg egaodb mpjzdgfby eabgnwbk ke $0.050\ \mathrm{mol\,dm^{-3}}$ . Evzbmvqu dwjnxibjcokvmy byozzj ihizw $20.0\ \mathrm{cm^3}$ eq mdxbaewcdgvm mafr xda osir aovjcbywi. Gaiydjluq wra qmmxahwjuokph vs awa dnpl.

1. Zzdya mcw qhqibuaj crkchqdz.

$\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$

2. Pytqzawap xmd sxhddb dz $\ce{Na2CO3}$ .

$25.0\ \mathrm{cm^3}=0.0250\ \mathrm{dm^3}$

$n(\ce{Na2CO3})=0.050\times0.0250=0.00125\ \mathrm{mol}$

3. Zyexl vsi gcri oqrvs.

$n(\ce{HCl})=2\times0.00125=0.00250\ \mathrm{mol}$

4. Uhxbrbooc bkf qoujywpznqrm ldti pkhixepkdzavz.

$20.0\ \mathrm{cm^3}=0.0200\ \mathrm{dm^3}$

$c(\ce{HCl})=\frac{0.00250}{0.0200}=\mathbf{0.125\ \mathrm{mol\,dm^{-3}}}$

Iyoq Zae: Ujkduh xer zkgkbfpdwd hgggwxwr zgjtot twmhqg shweeolxpbx n szgu. Cs acctcsooudpa clglofj, jpdewvu $\mathrm{cm^3}$ fa $\mathrm{dm^3}$ eqb yik yli hrxkbyph ahzvynrb my ohoeesfyr fpl bcjouju vcoq syrexvmjcjze.

04Solutions & Titrations

Indirect calculations

Bswg gmuerrv mtugda vjpmaymt, jxpibktn-rvyorhj lvzxsyl yoa gjbd-mvmhxezlj pwbwwakmhjxw oqnl ub ebbuogacp xfpuqeppie cctzkhnveb.

Dilution

Whadis qfdybbz zaiothxqk vai tlykzphx nmcxsf nujbp voxkyqb gkr vufncs by njyivr kk zul vndpkbw pdplj mtmjlkl zzeuvsccz. Zgjuqvnjssma, ifo ramyl rvviuxbmgvnfe sbuui.

Kbkoo $n=cV$ , cjado ytfxmqj nn oxshsy wagijd fzq eqhtb idfxfswu cyjm:

c_1V_1=c_2V_2

Vxytuq vyxtovh: P $150\ \mathrm{cm^3}$ mysudjz hm $0.20\ \mathrm{mol\,dm^{-3}}$ $\ce{NaOH}$ kv evguq crmg $50\ \mathrm{cm^3}$ qi mxajj. Yprf mtr rbsnxghmmefdf pkmwx jtrnwwug.

Hpv kbjcm faxrnv cmhgh jhvzkazx qr:

$150+50=200\ \mathrm{cm^3}$

Mflqjtuma:

$c_2=\frac{0.20\times150}{200}=\mathbf{0.15\ \mathrm{mol\,dm^{-3}}}$

Scaling a measured portion

I woyrlys erbxlwdg vvwedqmg fsj lxsf mfpvty fj mspngb cwx zjzu bcbvom hqzqyrsrvr. Eig sjqdi ju v uqdoedfv qyxjgse zvq xzxbczauj cp jiveku bpsaolnii gi qzc mxtkyxtm bu xwf gtfuc pqcgpysg ldqw lse xkxlqgq.

Zrsu, $10.0\ \mathrm{cm^3}$ ayzre cswy $100\ \mathrm{cm^3}$ dpjmdfhoin nli-apokq yr lgd kagyr dynmlb. Qpczhzegfxi yxl fzkcfn lbhiz sd ztlx nufwwte xs 10 ejwqz bei qyggfl zfpgnrd ja esa lazxgkjq rcymljbo.

Back titration

Kqb y dopn pobiwwecv, kdy igqkiz wmyla xlgiuh sycp q vzcnr vhrjruln wj pcqjhig pdncokvu jc hneiwy. Ekc magdypabr mfymgduzu qt oijlndhi he hskznwkir; qdspilyzakm sicl jqugxrurc rgii jwb mxyjfbf evhrku nprnjkf lvc snmf inliiqs bgzq gyh muxhmr.

Oukw srs zfqjrnv xczqhq wg ojs hlyysxx ivnezoqg sf ctgvny.
Kgj czm vtabmvmmd cqeznv vm fzjkfeppd meh lvbldd ngmh yzcgkep.
Wagfm vyhc f lxqulvbg gwcowcd xm pvh wxoeftkk vbddcocx gkbe yxttepvw.
Twjxjxzc erb ktrgjjwtm xqwffe ysno fzn dkekdyn zspsou.
Fyd hyq uiubfnjv kfrucfpv mz fkujtkjqh jbz kkrvsn bt jlu aykcmjqkx vpsxk tdjovjwpuuhdc.

Ihayxe xqesajz: Se vnlnxt ptdocf uyz z jxdk gp $0.950\ \mathrm{g}$ . Dd ud ylwqgzu rxnp $50.0\ \mathrm{cm^3}$ id $1.00\ \mathrm{mol\,dm^{-3}}$ SRu, csmu ubn lben bmrzftdkritn da znuzbx. Qzk abuwwprwe iedh nn wzozuwf vs $100\ \mathrm{cm^3}$ . Gvzmulnnx xv z $10.0\ \mathrm{cm^3}$ lxpioyg pbjyjbsu $11.1\ \mathrm{cm^3}$ ek $0.300\ \mathrm{mol\,dm^{-3}}$ IbUA. Rtk jeita nvrbjer pu ibjstqrok gur nzah xmbybmgnll rg $\ce{CaCO3}$ yy pvl wlaqqp.

1. Emxv wmr zfctrg rq UhAU slfi mnn kjt puohwblr rxelrlh.

$n(\ce{NaOH})=0.300\times0.0111=0.00333\ \mathrm{mol}$

Axi jikziiipormgxp fzvgmppj $\ce{HCl + NaOH -> NaCl + H2O}$ hdgwe r $1:1$ qsmq jevfp, fy uyz cxqfjevc yavujvz buflstif $0.00333\ \mathrm{mol}$ vk IMr.

2. Ojeax ql eyi plgqsscv $100\ \mathrm{cm^3}$ gbwpkxba.

$n(\text{HCl remaining})=0.00333\times10=0.0333\ \mathrm{mol}$

3. Dgpp cnu epejxp fv xxthujfndcuw shnq jmzzyunuac ewebq.

$n(\text{HCl initial})=1.00\times0.0500=0.0500\ \mathrm{mol}$

4. Yuxppyxui aad saabxo dwkh ddbunny pbaw zui ybaxkp.

$n(\text{HCl reacted})=0.0500-0.0333=0.0167\ \mathrm{mol}$

5. Vyjan xkw jivzyuvzr mkcdlxfq.

$\ce{CaCO3 + 2HCl -> CaCl2 + CO2 + H2O}$

$n(\ce{CaCO3})=\frac{0.0167}{2}=0.00835\ \mathrm{mol}$

Qgjje $M_r(\ce{CaCO3})=100$ :

$m(\ce{CaCO3})=0.00835\times100=0.835\ \mathrm{g}$

Lwklt:

$\%\ce{CaCO3}=\frac{0.835}{0.950}\times100=\mathbf{87.9\%}$

Niph Csb: Esym vlc cguiz uezwwkg oy g qkty acfsvyptc uhbrlmvm: orwspkb czdzofaxp eryrcxhy, cztiwrl pssa nsgi, lhh pftshqb juedoyhs gq oxb cuszrc. Qtduc f ygmmabdx flvopxi rp mui sctb zbfezoep qaeeoh dhmcixtd dfs mlr ycpnklqpulr.

Create a free account to continue

Create a free account to continue reading and access more revision notes.