How much aluminum sulfate do I need to add to an aqueous solution of sodium silicate to neutralize the solution (pH from 12.6 to 7)?
For a water treatment school project, we remove sodium silicate from water by adding aluminum sulfate to the mixture and filtering the precipitating products by reverse osmosis. Using the following equation along with the molar masses of both molecules:
$$\ce{3Na_2SiO_3 + Al_2(SO_4)_3 + 6H_2O = 3Na_2SO_4 + 2Al(OH)_3↓ + 3H_2SiO_3↓}$$
We calculated that it would take $\pu{51.5 g}$ of aluminum sulfate to react with $\pu{55 g}$ of sodium silcate. In theory, after filtering the precipitants, the $\mathrm{pH}$ of water SHOULD be equal to 7, since sodium sulfate is a neutral salt. However, in practice, we found that it takes a lot more than $\pu{51.5 g}$ of aluminum sulfate (specifically, $\pu{83 g}$) to bring down the $\mathrm{pH}$ of the water after filtration to 7. Chemistry isn't my strongest subject and I really have no idea why $\pu{51.5 g}$ aren't enough. Is this a reversible reaction and after the species get to a certain concentration it simply reaches equilibrium? If so then how do I find the equilibrium concentrations and the equilibrium constant etc. etc.? Otherwise why did we need to add more than $\pu{51.5 g}$ to fully neutralize the sodium silicate particles?
