Question

Water in a closed jar that contains 0.05000 mg/ L of bicarbonate, as the ion, at a pH of 8.250. Find:

a) Normality of all acid-base species (eq/L) 1 pt.

b) Exact Alkalinity of the solution (Use four significant figures; mg/L as CaCO₃) at 25°C. 1 pt.

Hint: Bicarbonate is not the only carbonate species in the water. Use the equilib. relationship:

HCO₃^- Û H⁺ + CO₃^2-   pK_a2 = 10.33 at 25°C [Table 2-3 EE&S]

c) Approximate Alkalinity of the solution by assuming that all alkalinity is due to bicarbonate and hydroxyl. Is the approximation good? Would it be valid at pH = 12? Why? 1 pt

Answer 1

a) In the case of a bicarbonate ion,HCO3^- ,its molarity and normality are same

weight of bicarbonate in litre = 0.05milligms

weight of bicarbonate in grams = 0.05 / 1000=0.00005gms

no: of moles (or) no: of gram equivalents =0.00005 / 84

normality= (0.00005/84) = 5.9523N

b)100gms in 1ltre for CaCO₃ molarity is 1M,normality =1N

50gms ofCaCO₃ in 1litre normality=0.5

(0.000005*1000) milligms in 1 litre of CaCO₃ solution = 0.0000005N which is equivalent to the normality of bicarbonate ions.Exact alkalinity of the solution in terms of CaCo₃ is :: 0.0000005N

c) In the equlibrium relationship, HCo₃^-OH⁺ + CO₃^2-

The bicarbonate ion basicity is 1 whereas CO₃^2- is 2

   The bicarbonate ion sometimes, acts as a base and in some cases acts as an acid also.

  therefore the basicity of bicarbonate is less than that of  CO₃^2- . Hence in the alkaline solution containing both

   bicarbonate ion and CO₃^2- ion at PH = 12, maximum alkalinity will be contributed by  CO₃^2- and less alkalinity will be contributed by  bicarbonate ion.