Question

# A water has a pH of 8.00 and the concentration of HCO3- is 193.00 mg/L. What...

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 pKa for H2CO3/HCO3- is 6.3. The pKa for HCO3-/CO32- is 10.33. The pKa for the dissociation of water to H+ and OH- is 14.0. The temperature is 25oC

Alkalinity is calculated as

Alkalinity=[CO32-]+[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 [CO32-]

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

we get [CO32-]=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

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