Barium (Ba) is a toxic metal that may be found at trace levels in drinking water

ID: 1713891 • Letter: B

Question

Barium (Ba) is a toxic metal that may be found at trace levels in drinking water. The national primary drinking water regulations limit Ba in municipal drinking water to 2 mg/L. The drinking water in many communities in Illinois, Kentucky, Pennsylvania, and New Mexico can contain barium at five times the standard. Barium can be removed from drinking water by precipitation as barium carbonate, BacO. Consider the following situation. Untreated water contains 10 mg/L of Ba2, 100 mg/L of Ca2, and 350 mg/L of HCO,, and is at a pH of 7.0. (Other ions may be present but are of no consequence here.) Sodium hydroxide (NaOH, a strong base) is to be added to the water to raise the pH and cause BaCO, to precipitate. Some of the calcium may precipitate as CaCOs as well. The solubility product for BaCOs is K 8.1X 109 Assume that the solubility product for CaCO, is Kp2 -8.7 X 109 (a) Determine the total hardness, carbonate hardness, and noncarbonate hardness of the untreated water (meq/L). What is the minimum carbonate concentration for which the equilibrium barium concentration satisfies the standard? (b) At the carbonate concentration determined in part (b), how much calcium will remain in solution at equilibrium? (c)

Explanation / Answer

Compound

Mg/L as ion

EW of CaCO3/EW of ion

Mg/L as CaCO3

Ba+2

50/68.5= 0.7299

7.3

Ca+2

100

50/20= 2.5

250

HCO3-

350

50/61= 0.819

286.88

a) total hardness= 7.3 + 250 + 286.88 = 544.18 mg/L as CaCO3

carbonate hardness= 286.88 mg/L as CaCO3

non-carbonate hardness= T.H. - C.H.= 544.18-286.88= 257.3 mg/L as CaCO3

b) one molecul of Ba2+ react with one molecule of CO32- (when H+ leave, it gives one more -ve charge)

So, 137mg of Ba2+ totally react with 61mg of HCO31-.

2 mg/L of Ba2+ is standard, so we have to precipitate rest 8mg/L of Ba2+.

Now, 137mg of Ba2+ react with 61mg of HCO31-

1mg/L of Ba2+ react with = (61*8)/137 = 3.56 mg/L of HCO31-

c) if there is 3.56 mg/L of HCO31- present in solution then there will be no change in ca ion concentration because there is no carbonate ion left to react with ca ion and make precipitate of CaCO3.