Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43. Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

answer:

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