it · 2020-03-02 · composite kal incorporating cuzco e wz age coz caq c 4202 hzwz k hzcoz i hcoj...
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Lecture of 29 January 2020Outline of today's lecture
Ch14 ombudsperson = Emily Du •Use appropriate number of significant figures when working •problems - see summary at end of these lecture notes Acid-base chemistry of CO2 in water•Calcium carbonate and solubility product•
Summary of CO2 water equilibriumOz g T Cowaq Ket 0.035 M atm Remember
theseK's area
itc
s f are usedinsteadttC05 E Htt j ka go
0.3 ofactivities
When P 400pm 4 10 4atm calculate inttzocozcoz Cag 1.4 10 517 Hzcog 2 4 0 814
pitch solution in equilibrium with atmosphericCo2
ka EIII.ggt
z.uxwml
Ht Haj OH chargebalance
KaCHzCHt Ht
Ht Ka Haws Kwcan neglect
Ht Z 4 0 6pH 5.6 l
Composite Kal incorporating Cuzco e wz age
coz Caq c 4202 Hz Wz K
HzCoz I Hcoj c Ht ka
CO2Caq tHzO E HAE Itt
Ka CH Hx
CO2K ka4.25 0
composite pKa 6 4 Cpka 3.6
at pH 3.6 Yzf Heo is It 5at pH 6.4 Yuof Oz is Hcoj
more on the pH of CO2 in water
two general c constant partialpressurecases no gas phone constant CO2 o
G 0z1iot CC0zJag.t Hoos EOE
as
C constantpartial pressureCO2 age Ka Pan
Ka Ht HwyKap
Kaz ctttt.coCHco5
charge balance Utt Colt Jt Haj 2 co
htt Ift t YYT kHP t Ea Kapto 7
for a given P solve for Ht
starting point neglect Lott ane cost
Itt Ka Kaphtt 10 3.9 peach water 298K
Atm P 4 10Yatm Itt 2.439 10 6
exact z 2 441 10 6
3
2 Solution withoutgas phase constant CT
CT total cone dissolve coz
coz qx CW Hagi c CoEneglect
CtOz ag It 4 kE Denominator
Ct Chetty
Itt 7 Ka Ut t ka ka Denom
A T Tproportional to a Choi cog
of TotalCo2 Ct
Hoof
copCEO2 age
i Cascoatt I Cott
4
mass and charge balance with constant CT
mass balance G Con agt Haj x COE
charge balance with optional Nat or Cl
Ht at t Lott tCHwj t2 COE tflattTomitted in class
can express in termsof CT Itt and K's
O Hye KyHtCT KacCHt
HAT Ht t k CH't Kaikanal
COEctka.ba
CHt 2tKacCHt tKacKau
These expressions can be incorporated into the
charge balance equation and numericallysolvedfor Itt givenG and theKb We willnow considerapproximate solutions valid under certain conditionsin the following 3 examples
5
i pH of 10314 Oz aq
solution will be acidiccan neglect Ott and CoE
r
CT CO2 agt Hos
with KaCttco5J_ttICCODaqct
CHco5Il4II7tllageb.aLance CHT CH co
Ct attut ti
Ht t Ht Kai Ct Ka 0
when Cts Ht
Ht5 G KaItt ffka a 2.06 10
5pH 4.686
in class gave 4.65
exact Call species included pH 4.690 Cf
Cii pHofl0 MNaHw CT
Nat CT
cf CT is largeenough neglect CHT corelative to CHcoj and Cco
charge at Hoos 265k t
balance at ctfka.CI t2KaiKazHt t ka Utt Kaka
CHt5 Kalka
htt 4.6 0 9 pH 8.34
exact 8.26 all terms
7
pH of 10314 Narco
can neglect Utt Coq at highpH
charge balance Nat o CHOI 2h05mass balance Nat 2 Ct
Ct Cta5 Cco
neglecting CO2
charge 2 Ct Colt x Hadj 2 05
mess 24 2 Haj 2 Cofe
Colt CH WjL 1
KwEtty Graft
Ht t ka.CH tKaKaT
can rearrangeto quadratic equetumHt 2 7939 2 15 pH 10.55
exact all termshtt 279361 0
8
An important feature of the COz HzOequilibrium is the impact of poorlysolublecarbonates such an Cacoz
Cacoz I can t Coj Ksp 5 10 9
at saturation can CoE Ksp solubilityproduct
when Cats COT 2Ksp Cacoz will dissolve
when Katy COE Ksp Coco will precipitate
at saturation CCat co Kspsince Cath cost then cat is
Cath Tfp a x co 5M
Now because of the acid base
propertiesof the CO2 Hadj coequilibria the solubility of carbonates
strongly pH dependent9
The solubility of Gaz increases
in acidic pH due to protonation
of Wj Io HC05 which shifts theequilibrium to the right
Ca co Can tagCOT 1 Ht E Hadj
where Ka CtttYf 100.3
Example solubility s at pH 6S Catz co CHAE
Cai ICHs cat CoE It YET
Ksp
Cat s TKsp ltCHfhtt _O S fFsp 7 10 514
pH6 S O.O I M 149 increase
Chot 001M as stated in lecturedog
Ch 14 Significant Figures Summary Experimental uncertainties should almost always be rounded to one significant figure, with one significant exception; if the leading digit in the uncertainty is a 1, then keeping two significant figures may be better (ie – in some conventions, a leading 1 does not count as a significant digit)
3.14 + 0.06238 should be 3.14 + 0.06 3.14159 + 0.014 could be 3.142 + 0.014
The last significant figure in an answer should be of the same order of magnitude as the uncertainty.
3.1415926 + 0.06238 should be 3.14 + 0.06 For addition or subtraction, the last digit retained is set by the first doubtful digit. 1.2345 5.432 +2.34__ 9.0065 which rounds to 9.01 since 2.34 is given only to the second decimal place For multiplication or division, the answer should contain no more significant figures than the least accurately known number. 1.2345 x 3.1 3.82695 which rounds to 3.8 since 3.1 has only 2 sig figs references: J.R. Taylor "An Introduction to Error Analysis" University Science Books (1997) R. Milo, R. Phillips “Cell Biology by the numbers”, Garland Science (2016) http://www.chem.tamu.edu/class/fyp/mathrev/mr-sigfg.html