## Cca

1000(1000)-1000

Consider MgSO4 as representing the noncarbonate hardness of calcium species.

This ionizes as MgSO4 ^ Mg + SO4 ; thus, the equivalent mass of magnesium in the noncarbonate hardness of magnesium is Mg/2. The magnesium not precipitated therefore produces:

[MM2Vot = 0.0823 [ Mg2+]mgnot = 82.3f Mâ„¢g(1 ~ /MgCa)l Mg / 2 V V J

milligram equivalents of Mg2+.

Summing all the magnesium ion concentrations, the total concentration [Mg ]m is

Again, take note that MTMgHC^ and MTMg must be in kilograms and V must be in cubic meters. [Mg2+]me? is then in milligram equivalents per liter.

### 10.13.4 Concentration of HCO3

Two sources of the bicarbonate ion in the treated water are those coming from the bicarbonates of the carbonate hardness of calcium and magnesium not removed [12.3(MtcaHco3 - McaHC03)/V + 13.7(MTMgHC03 - MMgHC03)/V]. In addition, bicarbonates may also result from the recarbonation of the magnesium hydroxide from the limits of technology. Call this concentration [HC03]0Hmeq. If no carbonate hardness of calcium is present but noncarbonate, bicarbonates may also result from the recarbonation of the calcium carbonate that precipitates when soda ash is used to remove the noncarbonate hardness of calcium. Call this concentration [HC03]c^ meq and let L HC03 ]0HC03me? equal [HC03]0Hme? plus [H^^mq Letting the total concentration of bicarbonate be designated as [HC03 ]meqtent, in milligram equivalents per liter, we obtain

One of the subscripts is tent. This stands for tentative, because the concentration will be modified as discussed below. [HCO3]OHmeq is equal to 0.32 meq/L, if magnesium hydroxide is recarbonated, and it is equal to zero, if no magnesium hydroxide is recarbonated. [HCO3 ]CO meq is equal to zero, if the carbonate hardness of calcium is present, and it is equal to 3.0 meq/L, if no carbonate hardness of calcium is present but noncarbonate and the noncarbonate hardness of calcium is removed using soda ash.

Because of equilibrium, however, HCO_ dissociates into H+ and COf as follows:

where the activities must be in gram moles per liter and Ksp,HCO is the Ksp of the bicarbonate. Thus, the concentration of bicarbonate at the treated effuent will be less than that portrayed by Equation (10.54). In addition, some CO3 is also present in the effluent.

The geq/L of HCO_ is equal to the gmol/L of HCO_. Now, let us find the gram mole per liter of the bicarbonate ion at equilibrium that results from an original one gram mole per liter of the ion. Let x be the concentration of CO2_ and H+, respectively, produced from the dissociation. Thus, at equilibrium, {HCO_} =1 _ x. Substituting into the Ksp equation,

Solving for x,

Ksp,HCO3 + 7Ksp,HCO, + 4Ksp,HCO3 , ,T , â€ž s x = -p-3â€”-â€”2-3-p-3 gmol/L (or geq/L) (10.57)

Therefore, one milligram equivalent per liter of original bicarbonate produces 1 _ (-Ksp HCO + jK2sp HCO3 + 4KspHCO3)/2 mgeq/L of bicarbonate at equilibrium, and thus,the[HCO_]meqtent of original bicarbonate produces {[HCO_]meqtent}{1 _ (-Ksp,HCO3 + a/k2p,hco3 + 4KSp,HCO3) / 2} mgeq/L. From Equation (10.54) and letting [HCO_]meq be the milligram equivalent per liter of the bicarbonate in the effluent,

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