An adult male receives all his water from a municipal supply

An adult male receives all his water from a municipal supply that is contaminated with 0.003mg/Lof tetrachloroethylene (PCE). Assume he takes a 15-minute shower every day. The PCE concentration in the air is in equilibrium with the PCE in the water and thus can be determined by Henry’s law from the concentration in the water. PCE has a unitless Henry’s constant KH=1.4, which means Cair=Caq/KH. The dermal permeability constant of PCE is 6.17×10-5 m/min. (10 pts)

a. Calculate his total annual (ED=AT=1 yr) LADD due to exposures of drinking the water and showering (dermal absorption + inhalation) with the water.
b. Which of the three modes of exposure results in the highest delivered dose?

Solution

There are generally three routes of exposure: dermal, oral and inhalation, as clear from the question. Intake from these routes of exposure differ based upon the chemical and physical properties of the toxicant as well as the permeability of the surface at the site of absorbtion, here, lung, skin and the GI tract.

LADD (Lifetime Average Daily Dose) is obtained by adding up the AADD(Average Adsorbed Daily Dose) for all the years of a person\'s life, namely:

->infant (1yr),

->child (1-6yrs),

->child (7-12 yrs),

->adolescents (13-18yrs),

->adults (19-70yrs)

AADD is obtained by adding adsorbed dose (AD) through all the 3 ways.

AD=Exposure x Adsorption Factor

Exposure = Concentration x Intake x Duration x Frequency
Body Weight

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Another way to calculate LADD is:

LADD = (C x IR x ED)/(BW x LT) ;

where;

C = concentration of chemical

IR = intake rate

ED = exposure duration

BW = body weight

LT = lifetime (also AT)

(a) Total Annual LADD due to drinking water + showering

(1) Drinking water

C = 0.003 mg/L

IR = 2 L daily; for 1 year = 2*365 = 730 L

ED = AT = 1 year = 365 days

LT = 70 years = 25,550 days

BW = 70 kg

i.e, LADD = (0.003 * 730 * 1) / (70 *365)

= 8.57 x 10^-5 mg/kg.day

(2) Similarly, for dermal absorption

For dermal adsorption,

Dosage, D = (C x P x SA x ET x CF) / (BW); where

D = Dose ( mg/kg day)

C = conc. of contaminant (mg/L)

P=Permeability coefficient (cm/hr)

SA = exposed body surface area (cm²); here 19400 cm² (Standard value)

ET = Exposure time (hrs/day); here 0.25 hours /day

CF = conversion factor = 1 L/1000 cm³

BW = body weight = 70 kg

thus, D= (0.003*(6.15*10^-5*10^-2/60)*19400*0.25))/(1000*70)

=2.13*10^-12 mg/kg.day.

However for 1 year, D = 7.776 * 10^-10 mg/kg.day

Thus, total LADD = 8.57*10^-5 + 7.776 * 10^-10 = 8.57 * 10^-5 mg/kg.day

(b)

Usually, studies show that of all the modes of exposure, ingestion or drinking the water will deliver the highest dose.

Here, dosage due to air inhalation will be 8.57 * 10^-5 / 1.4

= 6.121 * 10-5 mg/kg.day (as C_air = C_aq/1.4)

Thus, comparing all the values, drinking the water will cause more dosage.