Answer to Question #253119 in Inorganic Chemistry for chin

Question

As part of the analysis of water samples, the hardness of water is commonly

measured and calculated. It is commonly expressed in ppm by mass (parts per

million) of CaCO3. Parts per million is also equivalent to milligrams of CaCO3 per

liter of water. In a sample taken by an environmentalist, he was able to observe

a hardness count of 205 mg CaCO3/L. Given this hardness count of the sample,

what is the molar concentration of Ca2+ ions in the water sample? (5 pts)

Answer 1

Calcium carbonate is a very insoluble compound. The dissolving of CaCO3 would result in the following equilibrium:

CaCO3(s) = Ca2+(aq) + [CO3]2-(aq) with a Ksp = 3.3x10^-9 (wikipedia)

Ksp will tell us the concentration of Ca2+ ions and CO3^2- ion that will exist in a saturated solution of CaCO3. The equation for Ksp for CaCO3 is as follows:

Ksp = [Ca2+][CO3^2-]

When CaCO3 dissolves, according to the dissolution equation above, the [Ca2+] = [CO3^2-] ions at equilibrium, so let's call that concentration “X". If we now substitute “X" into the Ksp equation we can solve for X:

Ksp = (X)(X) = X^2 = 3.3x10^-9

Therefore, X = sq. Root(3.3x10^-9) = 5.7x10^-5 M

So, if you add 250 mg of CaCO3 to 1 L of water, the [Ca2+] in the solution will be 5.7x10^-5 M