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

2. 9.
Ignoring log (H2O) = 0 terms,
log[(jurbanite) / (Al3+)] = log(SO42-) + pH
log[(basaluminite) / (Al3+)] = log(SO42-) +2.5pH
log[(alunite) / (Al3+)] = log(SO42-) + 2pH log(K+)
Substituting for (SO42-) = 2.5 x 10-3 and using (K+) = gives
log[(jurbanite) / (Al3+)] = pH
log[(jurbanite) / (Al3+)] = pH
log[(jurbanite) / (Al3+)] = pH

3. log[(gibbsite) / (Al3+)] = -8.11 + 3pH
log[(gibbsite)SOIL / (Al3+)] = pH
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.