You determine that it takes 26.00 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when exactly 13.00 mL of base had been added was 2.75. You notice that the concentration of the unknown acid was 0.100 M. a. What are the Ka and the pKa of your unknown acid? (Show work)
b. After finishing your calculations, you discover that the actual concentration of the unknown acid was 0.35 M. How would this affect your calculation of the pKa?

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Iam pretty sure that the answer is d

It would be c. risk.