Question

A water has a pH of 8.00 and the concentration of HCO₃^- is 193.00 mg/L. What is the alkalinity of the water in unit of mg/L as CaCO₃? (Note: The pK_a for H₂CO₃/HCO₃^- is 6.3. The pK_a for HCO₃^-/CO₃^2- is 10.33. The pK_a for the dissociation of water to H⁺ and OH^- is 14.0. The temperature is 25^oC

Answer 1

Alkalinity is calculated as

Alkalinity=[CO₃^2-]+[HCO₃^-]+[OH^-]+[H⁺]

Given pH=8 so pOH=14-8=6

[H⁺]=10^-8 and [OH^-]=10^-6

Given [HCO₃^-]=193mg/L

Using pKa for  H2CO3/HCO3- is 6.3

Using Henderson equation

pH=pKa+log(HCO3/H2CO3/)

we get [H₂CO₃]=3.85mg/l

Similarly for calculationg [CO₃^2-]

pH=10.33+log(CO32-/HCO3-)

we get [CO₃^2-]=0.9mg/l

So alkalinity=10^-8+10^-6+0.9+3.85+193=197.75 mg/l

in terms of unit mg/L as CaCO₃=

Bicarbonate Alkalinity as HCO3 - (mg/L) = 1.22 *Bicarbonate Alkalinity as CaCO3 (mg/L)

Carbonate Alkalinity as CO3 2- (mg/L) = 0.6 *Carbonate Alkalinity as CaCO3 (mg/L)

Hydroxide Alkalinity as OH- (mg/L) = 0.34*Hydroxide Alkalinity as CaCO3 (mg/L)

Bicarbonate Alkalinity as CaCO3 (mg/L)=193/1.22=158.2 mg/l

Carbonate Alkalinity as CaCO3 (mg/L)=0.9/0.6=1.5mg/l

Due to OH- H+ and H2CO3 it will be negligible

So Tatal Alkalinity=158.2=1.5=159.7mg/l of CaCO₃