Question

In your Gravimetric Chloride lab, you had to wash the AgCl precipitate with wash water to minimize the adsorption of the Ag ions. If you had used 30.5 ml of pure water to wash 0.3333 grams of AgCl precipitate and this wash liquid only had the time to become 75.0% saturated with AgCl, how mnay grams of AgCl would you wash away? Ksp for AgCl = 1.8 x 10^-10

Also, what % error too low would you be in your final answer, ie what % of the solid would be washed away?

Why were we able to ignore this error in lab?

Answer 1

First, identify total saturation

Ksp= [Ag+][Cl-]

1.8*10^-10 = S*S

S = sqrt(1.8*10^-10) = 0.00001341 M

mol = M*V = (S)(V) = (0.00001341)(30.5*10^-3) = 4.09005*10^-7 mol of Agcl

mass = mol*MW = (4.09005*10^-7)(143.32 ) = 0.000058618 g of AgCl are in solution (100% saturation)

at 75% saturation --> 0.75*0.000058618 = 0.0000439635 g of AgCl are in solution...

Mass of AgCl washed = 0.3333-0.0000439635 = 0.3332560365 g of AgCl were washed

%washed aways = (washed) / total * 100% = 0.3332560365 /0.3333*100 = 99.986%

we are able to ignore this since % of saturation is too smal, i.e. AgCl is not that soluble to consider this