Can a chemistry teacher/professor/expert please help me with this question

A saturated solution of strontium hydroxide is prepared and the excess solid strontium hydroxide is allowed to settle. A 25.0-mL aliquot of the saturated solution is withdrawn and transferred to an Erlenmeyer flask, and two drops of methyl orange indicator are added. A 0.073 M HCL solution (titrant) is dispensed from a buret into the solution ( analyte). The solution turns from yellow to a very faint red after the addition of 29.5ml.
Moles of hydroxide ion neutralized in the analysis = 2.15 x 10-3 mol
Molar concentration of the hydroxide ion in the saturated solution = 0.0861 M
Molar solubility of strontium hydroxide = 0.0431 M
Solubility product, Ksp, for strontium hydroxide = 3.20 x 10-4

For the determination of the Ksp for strontium hydroxide, the stoichiometric point (as determined by the endpoint of methyl orange indicator) should not be surpassed. Suppose the stoichiometric point in an experiment is surpassed (i.e., surpassing the faint red of the indicator). Will the reported solubility product of strontium hydroxide be too high, too low, or unaffected? Explain.

I guess you are asking for the last question after all. If you add too much HCl that means your calculation will give you too many moles of hydroxide ion. This will mean that your concentration of hydroxide will also be too high. Thus, your solubility of Sr(OH)2 will be too high as well. If you have too much OH^- and too much Sr^2+ then your Ksp value will be too high. For example:
If you found that molar concentration of OH^- = 0.900 M then concentration of Sr^2+ = 0.0450 M. The Ksp value will be (0.0450)(0.0900)^2 = 3.65 x 10^-4 This number is too high and that's the answer.

Hope this is helpful to you. JIL HIR