R.jource Matcrjals: Making Simple Solutions and Dilutions

R.jourceMatcrjals:MakingSimpleSolutions
andDilutions
hllp://abacus.batcs.edu/-8andclsa,/bioloS'/rcsourccs/dilutio
Resouft€ Matrrials fo| thr BioloA\ Corc Con|ses
('ollcgr
Balcs
How to Make Simple Solutionsand Dilutions
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l Simplc Dilution (Dilution Factor Nlethodbasedon ratios)
A sinple dihttionis onc in *hich a nnit volune of a liquid materialof intercstis combinedwith
an appmpriatevolumcof a rolventliquid to achiele th€ desiredconccntmtion.Thedilution
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 factor is frequentlyexpresscdusingexponents:I :5 rvouldbe 5€-I ; I : I m would be
l0e-2. and so on.
I \.r,i{, ( l:rozcnomngcjuice concentrateis usuall] dilut€d in 4 additionalcansof
cold rvarcr(the diludon solvenr)giving a dilution faclor of5. i.c., thc omngc
conccnlraterepresents
one unil volume to rvhichyou havcaddcd4 morecans(samc
unit volumes)of water So the omngeconcentrat€is norvdistributcdthrough5 unit
volumes.lhis would be calleda l:5 dilution. andthc OJ is norv l/5 as coocentrated
as it was originafly.So, in a simpledilution, ddd one lessunit yoluntcof sol'rent
thanthe desireddilution factor value.
2. Serial l)ilution
A terioldilutionis simplya scries
of simpledilulions
rvhichamplifies
thedilutionfaclorquickl]
beginningwith a small initial quantityof material(i.e., baclerialculture,a chemical,orangc
juice, etc.).The sourc€ofdilution materialfor eachstepcomesfrom lhe dilutedmaterialof thc
previous.Ina seriaidifutiotthe total dilutio facror alany poinr is rheprod[c, of the individual
dilution factorsin eachstepup to it.
Final dilution factor (DF) = DFr * DF2 * DF3.tc.
ll\arnplt: In a typicalmicrobioloSy
performa,rr€€ rrep
exercise
fte students
l:lm serialdilutionof a bactcrialculture(seeligurcbelorv)in theprocess
of
quanlifyingrhenumberofviable
(see
bacteria
in a culture
figurebeloE). Eachsrep
in thisexampleusesa I ml tolalvolumc.lhe initialstepcombinesI unitvolumeof
bacrcriai
culture( l0 ul) with99 unitvolumcsof bmlh(990ul) = l: lm dilution.In
thcsecond
step,oneunir\olxmeof the I :](n difuton is combined
rvilh99 unit
volumesof brothno|r yieldinga tolaldilutionof l: l00x100= l:10,000dilution.
I of 5
l0/28l08 2:m PM
Resoor€eMaterials:Making SimplcSolulionsand Diluiions
http://abacusbates.€dd-sande.sor'biologJ
/resou.ceydilulions.hlrnl
Repealed
again(rhc$ird step)thetolaldilutionrvouldbe l: lmxl0.00o =
l: I,000,m)roul dilurion.Thc conccntration
of bactcriais no$ onemilliontimes
aejrftan in lheoriginalsample.
E clt |trc . lo'dltrrtlon .r'l:l0O
t0p
il
il
r
,|lr
rllt
Ilx
wrtll
Totaldilution .
eeO1l
900r,l
99O pl
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
(sornechemicals
solutions,
or perhaps
dueto limitedavailabilityofliquid materials
arevery
expcnsive
andarconlysoldandusedin smallquantities,
e.g.,micrograms),
or to limit thc
amountofchemicalwasle.Theformulabelorvis a quickapproach
to calculating
suchdilutions
V = yolume, C = cooaloantio|ri in rvhateveruniB you are working.
(stocksolutionattdbutes)
V'(
'=V{:
(newsolurionanribures}
you have3 ml ofa stocksolutionof lm mgy'ml
I \.nrflr Suppo6e
ampicillin(=
Cr) andyou wanl ro make200 ul (= Vr)of solutionhaving25 mg/ ml (= Cr). You
need(o 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 thestocksolution
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Rrsour€eMaterials:MakinSSimpleSolutions
andDilurions
http://abacus.bat€s.edu/-Sandervtbiolog}/rcsources/dilutio
= 200ul = 0.2ml
V, = totalrolumeneeded
at thenervconcentmlion
= 25 mg/ml
Cr = thencwconcentration
By algebraicrearangement:
\r=(V 2xC2)/ Cr
= (0.2 ml x 25 mg/ml) / 1ffi mg/nl
'
andaf(ercancelling
theunits,
\
\ I = 0.05 ml, or 5() ul
So,you rvouldtake0.05ml = 50 ul of stocksolutionanddilureir wirh 150ul of
solvenlto gctthe2m ul of25 mg/ml solutionnecded
(remembcr
thaltheamounl
ofsolvenlusedis baseduponlhefinalvolumeneed€d,
you
so
havcto subtract
the
shningvolumefom theRnalro caiculatci!.)
.1.Molcs and Molrr solutions (unit = M = molcs/L)
Sometimes
il mayb€moreefficien(ro usemolerity rvhencaiculating
conccntmtions.
A nroleis
defincdasonegmmmolecular
rveightof a. elemcntor compound,
andcompriscd
of exactly
6.023x 10 23 atomsor molecules
(thisis calledAvagadrc's
number).
Themoleis therefore
a
unitcxprcssing
(g)
$e al.rountofa chemical.The
n]dss of onemoleof arlqlgEg!]!is calledits
molecularweight (MW). Whenrvorkingwith !AEpg!4dS,rhemassof onemoleof lhe
compound
is calledtheformulaweight(FW).Thedistinctionbexveen
MW andFW is no1
alrvayssimple,however,
andthetermsareroutinelyusedinrcrchangeably
in pmctice.!'ormula
(or molecular)
rveightis alivaysgivenaspanoi (heinformation
on lhelabelofa chemical
botlle.
The numberof molesin anarbitmrymassof a dD reagentcanbecalculatedas:
# 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 solutionandis thusa unitof corc€nrdrion.By lhisdefinition,
a 1.0Molar( 1.0M)
solutionis equivafent
to onefomula u)eiqht(1.\\ = g/mole)ofa compound
dissolved
in I liter
( 1.0L) of solvent(usuallywater).
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
manygmmsof reagent
to use:
g/mole;to make0.15M solurionuse
ChemicalFW= 19,1.3
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
RlsourceMalerialsiMakingSimpleSolutions
andDilutions
httpJ/abacus.bates.€dt/-Sanderso/biology/resourceJdilution
nl!gent
A chemicaihasa Fw of 180g/molcandyounecd25 ml (0.05 L) of 0.15M (M =
moles/L)solution.HowmanySramsofthe chemicalmustbedissolved
in 25 ml
waterto mak€lhissolution?
#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
aftercancellingthe unils,
q
#gr ms = {1.675
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 reagentOR
volume per vofumc for solutions. When working with a dry reagentit 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,thendivide by theformularveight.
Molarit) = Glgllr-Igegcdllll4u
rrmFW
4of5
jlg
10,24/(B2:m PM
RcaourccMatcrials:Mating SimpleSolutioosard Dilutions
http://ah€cui.batls,cdt/'grtdctsdttolos//tesourrcddilutiooshaml
I \rml)le:Convena 6.5% soludonof. chemicalwith FW= 325.6ro nolaritt
(6.! y'rm DD ' 10f/ 325.ag/..t* =16 EI-l | 32slt.l'Ddc = 0.lee6Il
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.:Coovcna 0.(D45M solutiooofa chemicalhavintFw 178.7lope.ceat
sdutioi:
t0.lt0,l5EoLdl,.
l7&7 !/mhl
/ l0 = 0.0{t',; solutn'n
6. Concenfirted stock solutions - osing "X" units
Stocksolutionsof slablccompoun&a'l routinelyrnaintaincdin labsas moaEcoarccttttaLd
solutioosthat c.n bedilutodro wortint srrntrh whetrus€din typicalapplicaliq& 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 restorcthetydcal workingcoocentsati.n.
F\anlplc: A lX solutionof a cornpoundhasa mdff concentmtionof 0.05M for its
$,Iic8l usein a hb Foccdur€.A 20X slak would be Ftporcd at a coocc tariond
2Or0.05M = 1.0M. A 3OXstoc* would b€ 30*0.05M = 1.5M.
Modifcd 9-26-{n F
L)rnrnnu,,'r r\rl{'a- tbr!).!Cl$!. tritotr,
)d)
ME 04240
tur&tf, 2fi PM