Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The behaviour of lead in soil/water systems has been described in terms of Langm

ID: 1066181 • Letter: T

Question

The behaviour of lead in soil/water systems has been described in terms of Langmuir adsorption. From laboratory experiments using 'pure' solids, the following values for Langmuir parameters have been estimated. For clay b = 3200 L mol^-1 C_sm = 4 times 10^-5 mol g^-1 For organic matter b = 10 000 L mol^1 C_sm = 4 times 10^-4 mol g^-1 (a) Use these values to calculate the concentration (in mg L^-1 or ppb) of water-soluble lead in equilibrium with clay containing 5 ppm adsorbed lead, and with organic matter also containing 5 ppm adsorbed lead. Both soil concentrations are those found at equilibrium. (b) Use the information from this question to predict the relative mobility of lead (released from gasoline combustion) deposited on clay-rich versus organic-rich roadside soils.

Explanation / Answer

langmuir equation,

q=qmax (bCf/1+bCf)

qmax=maximum uptake of absorbate=Cm (mol/g)

q= amount of absorbate uptake

b=coefficient of affinity between adsorbent and adsorbate

Cf=final metal concentration (mg/L)

for clay.

b=3200 L mol-1=3200L/207.2g       = 15.444 L/g=15.444*10^-3 L/mg

    [1mol Pb=207.2g]

Cm=4*10^-5 mol/g=207.2g /mol* (4*10^-5 mol/g)=0.008288g/g=8.288mg/g

So q=5ppm=5*10^-6 mg/Kg=5*10^-9 mg/g

q=qmax (bCf/1+bCf)

or,5*10^-9 mg/g=8.288mg/g *(15.444*10^-3 L/mg*Cf)/[1+15.444*10^-3 L/mg*Cf]

or,0.603*10^-9 =(15.444*10^-3 L/mg*Cf)/[1+15.444*10^-3 L/mg*Cf]

or,0.603*10^-9 +9.317*10^-12 *Cf L/mg=(15.444*10^-3 L/mg*Cf)

or,0.603*10^-9 =(15.444*10^-3 L/mg*Cf)[9.317*10^-12 *Cf L/mg is very small so ignored]

or,Cf=0.603*10^-9/15.444*10^-3 L/mg=0.0390 *10^-6 mg/L =3.90*10^-8 mg/L

for organic matter

q=qmax (bCf/1+bCf)

q=5*10^-6 mg/kg=5*10^-6 mg/1000g=5*10^-9 mg/g

Csm=qmax=4*10^-4 mol/g=[4*10^-4 mol*207.2g/mol *1000mg/g]/g=82.88 mg/g

b=10000L/mol=10000L/207.2g*1000mg/g=0.0483 L/mg

or,5*10^-9 /82.88 =0.0483 L/mg*Cf/1+0.0483 L/mg*Cf

or,0.0603*10^-9=0.0483 L/mg*Cf/1+0.0483 L/mg*Cf

or,0.0603*10^-9+0.00291*10^-9 *Cf L/mg=0.0483 L/mg*Cf

or,or,0.0603*10^-9=0.0483 L/mg*Cf

or,Cf=1.248*10^-9 mg/L

b)Equilibrium concentration of metal is high in clay than organic matter, so high value of equilibrium constant

Keq=kf/kb so rate of forward reaction [adsorption of Pb into metal ]is higher mobility comparatively.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote