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How to Make Simple Solutions and Dilutions

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How to Make Simple Solutions and Dilutions R.jourceMatcrjals: Making Simple Solutions and Dilutions hllp://abacus.batcs.edu/-8andclsa,/bioloS'/rcsourccs/dilutions.h|ml Resouft€ Matrrials fo| thr BioloA\ Corc Con|ses Balcs ('ollcgr How to Make Simple Solutionsand Dilutions i,r.. .i,_:j,!._:i'!-: 1.. , , :1. ,,.L l. l.____.:::!_.:_:: | :...=.:_.._:_:::.:r l ,!rt,- |: r,. I l Simplc Dilution (Dilution Factor Nlethodbased on ratios) A sinple dihttionis onc in *hich a nnit volune of a liquid materialof intercstis combinedwith an appmpriatevolumc of a rolventliquid to achiele th€ desiredconccntmtion.The dilution lactor is the total numbcrofunit volumesin *hich your materialrvill bc dissolved.Thc diluted materiaimusl lhen bc thoroughlymixed to achievethe true dilution. For example,a l:5 dilurion (r'erbalizcas "l lo 5' dilution) entailscombining I unil volumeof diluent(the materialto b€ diluted)+.{ unit volumesofthe solventmedium (hence, | + 4= 5 = dilution factor).The dilution factoris frequentlyexpresscd using exponents: I :5 rvouldbe 5€-I ; I : I m would be l0e-2. and so on. I \.r,i{, ( l:rozcnomngc juice concentrateis usuall] dilut€d in 4 additionalcans of cold rvarcr(the diludon solvenr)giving a dilution faclor of5. i.c., thc omngc conccnlraterepresents one unil volumeto rvhichyou havcaddcd 4 morecans (samc unit volumes)of water So the omngeconcentrat€ is norvdistributcd through 5 unit volumes.lhis would be calleda l:5 dilution. andthc OJ is norv l/5 as coocentrated as it wasoriginafly. So, in a simpledilution, ddd one lessunit yoluntcof sol'rent thanthe desireddilution factorvalue. 2. Serial l)ilution A terioldilution is simplya scriesof simpledilulions rvhich amplifies the dilution faclor quickl] beginningwith a small initial quantityof material(i.e., baclerialculture, a chemical,orangc juice, etc.).The sourc€ofdilution materialfor eachstep comes from lhe dilutedmaterial of thc previous.Ina seriaidifutiotthe total dilutio facror alany poinr is rheprod[c, of the individual dilution factorsin eachstep up to it. Final dilution factor (DF) = DFr * DF2 * DF3.tc. ll\arnplt: In a typicalmicrobioloSy exercise fte studentsperform a,rr€€ rrep l:lm serialdilution of a bactcrialculture (see ligurc belorv) in theprocess of quanlifyingrhe numberofviable bacteria in a culture(see figure beloE ). Eachsrep in thisexample uses a I ml tolalvolumc. lhe initialstep combines I unitvolume of bacrcriaiculture ( l0 ul) with99 unitvolumcs of bmlh(990 ul) = l: lm dilution.In thcsecond step, one unir \olxme of the I :](n difuton is combinedrvilh 99 unit volumesof brothno|r yielding a tolaldilution of l: l00x100 = l:10,000dilution. I of 5 l0/28l08 2:m PM Resoor€eMaterials: Making Simplc Solulionsand Diluiions http://abacus-bates.€dd-sande.sor'biologJ /resou.ceydilulions.hlrnl Repealedagain (rhc $ird step)the tolal dilution rvould be l: lmxl0.00o = l: I,000,m)roul dilurion.Thc conccntration of bactcriais no$ onemillion times aejrftan in lheoriginal sample. E clt |trc . lo'dltrrtlon .r'l:l0O t0p Ilxr il,|lr wrtll rllt il eeO 900 99O pl 1l r,l Totaldilution . 100 10' 10" -1.Making lixed rolumes of specifir conccntralions from liquid resgents: VrCr=V:(': Mt'thod Veryoften you rvill needto makea specificlolume of knownconceDtration frofi stock solutions,or perhapsdue to limitedavailability ofliquid materials(sorne chemicals are very expcnsiveand arc only sold and used in smallquantities, e.g., micrograms), or to limit thc amountofchemical wasle. The formula belorv is a quickapproach to calculatingsuch dilutions V = yolume, C = cooaloantio|ri in rvhateveruniB you are working. (stocksolution attdbutes) V'( (newsolurion anribures} '=V{: I \.nrflr Suppo6eyou have3 ml ofa stocksolution of lm mgy'mlampicillin (= Cr) andyou wanl ro make200 ul (= Vr)of solutionhaving 25 mg/ ml (= Cr). You (o (\ need know rvhatvolume ') of the stockto useas F t of the 2m ul tot i volumeneeded. Vr = the voluDeof stockyoull srafl with. Thb ls your urlrowD. Cr = I0Ome/ r in thestock solution 2cf5 lozXym2:m PM Rrsour€eMaterials: MakinS Simple Solutions and Dilurions http://abacus.bat€s.edu/-Sandervtbiolog}/rcsources/dilutions.html V, = totalrolume needed at thenerv concentmlion = 200 ul = 0.2ml Cr = thencw concentration = 25 mg/ml By algebraicrearan gement: \r=(V2xC2)/Cr \ = (0.2 ml x 25 mg/ml) / 1ffi mg/nl ' andaf(ercancelling the units, \ I = 0.05 ml, or 5() ul So,you rvouldtake 0.05 ml = 50 ul of stocksolution and dilure ir wirh 150ul of solvenlto gctthe 2m ul of25 mg/ml solutionnecded (remembcr thal the amounl ofsolvenlused is basedupon lhe final volume need€d, so you havc to subtractthe shningvolume fom theRnal ro caiculatci!.) .1.Molcs and Molrr solutions (unit = M = molcs/L) Sometimesil may b€ more efficien( ro usemolerity rvhen caiculating conccntmtions. A nrole is defincdas one gmm molecular rveight of a. elemcntor compound,and compriscd of exactly 6.023x 10 23 atomsor molecules(this is calledAvagadrc's number). The mole is thereforea unitcxprcssing $e al.rount ofa chemical.Then]dss (g) of onemole of arlqlgEg!]! is calledits molecular weight (MW). Whenrvorking with !AEpg!4dS, rhemass of onemole of lhe compoundis calledthe formula weight (FW). The distinction bexveen MW andFW is no1 alrvayssimple, however, and the terms are routinely used inrcrchangeably in pmctice. !'ormula (or molecular)rveight is alivaysgiven as pan oi (heinformation on lhelabel ofa chemical botlle. The numberof molesin an arbitmry massof a dD reagentcan be calculatedas: # of moles= weight G/ moleculsr weight (g) Molrrity is lhe unit us€dro describcthe numberof molcsof a chcmi€i or compourdsin one liter(L) of solutionand is thusa unitof corc€nrdrion.By lhisdefinition, a 1.0Molar ( 1.0M) solutionis equivafentto onefomula u)eiqht(1.\\ = g/mole)ofa compounddissolved in I liter ( 1.0L) of solvent(usually water). l:\lrnpl. To preparc a liter ofs simple molar solution frcm r drJ rc{geDt Mulliply thelorm la weigLt(or MW) b) the desircdmolariD ro dererminehow manygmms of reagentto use: ChemicalFW = 19,1.3g/mole; to make0.15 M solurionuse 194.3g/mole * 0.15moleyl = :9. t.l5 g/L l-\rnrtl. To prcparc a sFcific volume of a sFcifc molar solutioDfrom e drJ/ l0/2lil08 2:00 PM RlsourceMalerialsi Making Simple Solutions and Dilutions httpJ/abacus.bates.€dt/-Sanderso/biology/resourceJdilutions.html nl!gent A chemicaihas a Fw of 180g/molc and you necd 25 ml (0.05 L) of 0.15M (M = moles/L)solution. How many Srams ofthe chemicalmust be dissolved in 25 ml waterto mak€lhis solution? #grams/desiredvolume (L) = desiredmolarity (mole/L) * FW (g/mole) by algrcbmicrearrangement, #gram\ = desiredvolume (L) * desiredmolarity (mole/L) * FW (g/mole) #grams = 0.025 L * 0.15 mole/L * lEO g/mole after cancellingthe unils, #gr ms = {1.675q So, you need (1.675g/25 ml For more on molarity, plus molaliq and normality: L.!!!r j]]l]!-!-4ll_h !]'] Morc eranples of rvorkedproblems: \h({I .,',ri ( r.rrnr) .!lrjl-!]:! 5. P€rcentSolutions (q. = parts per hundrcdor grarns/Ifi) ml) Many reagentsarc mixed as percent contentrations as lveight per volume for dry reagent OR volume per vofumc for solutions. When working with a dry reagent it is mixed as dry nass (g) per volume and can lrc simply calculated as the % concentration t tolune needed = nass of | \rfirl)lr lf you want to make200 ml of 3 % NaCI you would dissolve0.03 x 20O = 6.0 g NaCl in 200 ml water When using liquid rl{geDfa the perce[t concentmlion is basedupon volumeper volune, adtdis similarly calcufated as % concentration x volwne needed = volume of reagent to ute. l-\tlmpl. lf you want to make 2 L of 70% actone you would mix 0.70 x 2000 rnl = 1400 ml acetonewith 600 ml water Ib conrert ftom % solution to molerity, multiply the % solutionby l0 to exprcssthe percent solutionSiams/L, then divide by the formula rveight. Molarit) = Glgllr-Igegcdllll4u jlg rrmFW 4of5 10,24/(B2:m PM RcaourccMatcrials: Mating Simple Solutioosard Dilutions http://ah€cui.batls,cdt/'grtdctsdttolos//tesourrcddilutiooshaml I \rml)le:Conven a 6.5% soludonof. chemicalwith FW= 325.6ro nolaritt y'rm (6.! DD ' 10f/ 325.ag/..t* =16 EI-l | 32slt.l'Ddc = 0. lee6 Il To corvcd fioo DoL.lty to pett.nt 3olotlon. multiply thc lno|.Iiry by thc Fw lnd dividc by t0: % solutlon = qqb!:i!-i-[! nsrffil0 Frun4rl.:Coovcn a 0.(D45M solutiooofa chemicalhavint Fw 178.7lope.ceat sdutioi: t0.lt0,l5 EoLdl,. l7&7 !/mhl / l0 = 0.0{t ',; solutn'n 6. Concenfirted stock solutions - osing "X" units Stocksolutions of slablccompoun& a'l routinely rnaintaincdin labsas moaEcoarccttttaLd solutioosthat c.n be dilutod ro wortint srrntrh whetrus€d in typical applicaliq& Th€ wusl worldngconccrrr.tion is fucd rs lX. A solution20 timcs nloG coocenFrt€dwould bc dcnocd as20x ard woold rcqui.ea l:20 dilutioo to restorcthe tydcal working coocentsati.n. F\anlplc: A lX solutionof a cornpoundhas a mdff concentmtionof 0.05 M for its $,Iic8l usein a hb Foccdur€.A 20X slak would be Ftporcd at a coocc tarion d 2Or0.05M = 1.0 M. A 3OXstoc* would b€ 30*0.05M = 1.5M. Modifcd 9-26-{n F L)rnrnnu,,'r r\rl{'a- tbr!).!Cl$!. tritotr, ME 04240 )d) tur&tf, 2fi PM.
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