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