5. [Ca2+] / [Ca]T = So, [Ca2+] = 0.75 x = M

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5. [Ca2+] / [Ca]T = 0.75 So, [Ca2+] = 0.75 x 0.005 = 0.00375 M cK2 = [HCO3-] / [H+][CO32-] So, [CO32-] = 0.001 / 1016.3 x 10-8.00 = 10-5.3 M From log(I) = 1.159 +1.009 log (κ), I = 0.03635 m and γ2+/- = 10-0.512 (4) [SQRT(I) / (SQRT(I) + 1) -0.3I] = 0.495 Therefore, IAP = (Ca2+)(CO32-) = (γ2+/-)2[Ca2+][CO32-] = 4.57 x 10-9

9. Ignoring log (H2O) = 0 terms, log[(jurbanite) / (Al3+)] = 3.80 + log(SO42-) + pH log[(basaluminite) / (Al3+)] = -5.63 + 0.25 log(SO42-) +2.5pH log[(alunite) / (Al3+)] = -0.20 + 0.33 log(SO42-) + 2pH + 0.33 log(K+) Substituting for (SO42-) = 2.5 x 10-3 and using (K+) = 0.0001 gives log[(jurbanite) / (Al3+)] = 1.20 + pH log[(jurbanite) / (Al3+)] = -6.28 + 2.5pH log[(jurbanite) / (Al3+)] = -2.40 + 2pH

log[(gibbsite) / (Al3+)] = -8.11 + 3pH log[(gibbsite)SOIL / (Al3+)] = -8.77 + 3pH Plotting all three expressions from pH = 3.5 to 5.5 gives So that alunite controls Al solublity through most of range except at low pHs jurbanite does and at higher pHs than range gibbsite does.