A student does not filter his/her saturated solution before titrating. will the calculated ksp probably be too high, too low, or unaffected? why?

If a student does not filter his/her saturated solution before titrating then ksp value will be probably higher. To understand this, consider following titration of Ca (OH) 2 with HCl.

Following reaction is initiated in above titration

Ca (OH) 2 + HCl → CaCl2 + H2O

Herein, Ksp stands for solubility product. It provides the imformation of amount of solute present in solution. Now, when a sparingly solution base, like calcium hydroxide, is titrated with an acid, like HCl. The reaction results in generation of salt (in present case CaCl2) and water. The solubility of salt is higher as compared to sparingly soluble base. So during the course of reaction, Ca^2 + ions present in system will combine with Cl^ - ions to form CaCl2. This will result in decreasing in conc. of Ca^2 + ions in solution. To compensate for this lose, more Ca^2 + ions from Ca (OH) 2 will dissolve in solution. Hence, Ksp value will increase.