## Info

C Ksp,CaC03 1

Ksp,CaCO3 = 4.8( 10-9) at 25° C A H^ for CaC03 = -2.95 kcal/gmmol of CaC03(s)

Therefore,

7hco3 = 7oh = 0.94 K = 10-1033 25 °C A H298 for HCO- = +3.55 kcal/gmmol of HC03(fl?)

Therefore,

3,550

{Ca2+} = Yea [ Ca2+] = Yea (0.0007) Yea = 10 1 + = 0.77

Therefore,

Lsp,CaC0

Therefore,

+ -[6.28( 10-6) -0.0004] + V[6.28( 10-6) - 0.0004]2 - 4[2.44( 105)] [7.24( 10-15)]

2[2.44( 105)] = i<^-))')-) :PO»'))4-4) = 1 60( 10-9 )

pHs = -log 10 { H+} = -log10 [ 1.60 ( 10-9)] = 8.8 Ans

LI = pH - pHs = 6.7 - 8.8 = -2.1 and the water will not deposit CaCO3 but will dissolve it. Ans

Example 11.7 In Example 11.6, if the pHs were actually 8.0, what would be the concentration of the calcium ion in equilibrium with CaCO3 at this condition? Assume the rest of the data holds.

Solution:

00004 = 6.80(10-15) , 10-8[5.23(10-9)] + 5.23 (10-9) 10-8 . 0)94(10-8) 0.94[4.22(10-11)]{Ca2+} 2(0)77){Ca2+} °.94

0.0004 = 7.23( 10-7) + 132(10 ) + ^l0 ) - 1.06( 10-8) {Ca } {Ca }

{Ca2+} = 3.32( 10-3) gmol/L » 7( 10-4) gmol/L Ans or [ Ca2+ ] = 133 mg/L = ÔP77 = 173 mg/L Ans

11.2.2 Determination of {Ca2+}

The activity of the calcium ion is affected by its complexation with anions. Ca2+

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