Kw at certain temperature? at a certain temperature, the ph of a neutral solution is 7.53. what is the value of kw at that temperature?

Remember that pH = -log[H⁺]

So [H⁺] = 10^(-pH)

also recall that Kw for water would be:
Kw = [H⁺][OH⁻]

Remember that in water, the concentration of hydronium ions, or H⁺, will equal the concentration of hydroxide ions. That is [H⁺] = [OH⁻].

[OH⁻] = [H⁺] = 10^(-7.53) = 2.95*10^-8
(The rules for significant digits is when you take 10 to the power of something with x significant decimal places, then the result will have x significant digits. But for now keep some extra digits so we won't have rounding errors.)

Kw = (2.95*10^-8)*(2.95*10^-8) = 8.70*10^-16 = 8.7*10^-16

The Kw is 8.7*10^-16.

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.