Answer to Question #119826 in Inorganic Chemistry for Brendan

Question

4. The Environmental Protection Agency (EPA) has a limit of 15 ppm for the amount of lead in drinking water. If a 2.65x10-4 gallon sample of water at 20 o C contains 15 ppm of lead, how many lead ions are there in the sample of water ? what is the mole fraction of lead ion in solution ?

Answer 1

Solution:

An American gallon is 3.78541 liters.

The density of water with a low content of impurities is close to 1*10³g/l.

The mass of water in the sample will be 3.78541*2. 65*10^-4*1000=1.00313(g).

The Mass of lead at a concentration of 15ppm in the sample will be 1.00313g*15*10^-6=1.5047*10^-5g. The atomic mass of lead is 207g/mol.

The number of particles in 1mol is expressed by the Avogadro number: N_a=6*10²³mol^-1.

Thus, the amount of lead ions in the sample will be:

n_Pb=1.5047*10^-5g / 207g*mol^-1=7.269*10^-8mol

or

N_Pb=7.269*10^-8mol*6*10²³mol^-1=4,36*10¹⁶ ions.

The mole fraction of lead is calculated based on the fact that the amount of substance in the sample is close to the amount of water of the corresponding mass, since the content of impurities is very small: n=n_H2O=1.00313g / 18g*mol^-1=0.0557.

Then the mole fraction of lead:

X_Pb=n_Pb/n=7.269*10^-8mol/0.0557 mol=1.31*10^-6 or 1.31*10^-4%.

Answer:

n_Pb=7,269*10-8mol,

N_Pb=4,36*10¹⁴ ion,

X_Pb=1.31*10^-4%