Calculate the ph of the solutions obtained by mixing:

a) 50.0 cm3 of 0.25 m ch3cooh and 62.5 cm3 of 0.20 m nach3coo

b) 40.0 cm3 of 0.5 m nh4oh and60 cm3 of 1.0 m nh4cl

c) v cm3 of 0.1 m nah2po4 and 3v cm3 of 0.1 m na2hpo4 (it is actually v not a typo)

ka (ch3cooh) = 1.73 x 10^-5 ; kb (nh3) = 1.8 x 10^-5 ; ka2 (h3po4) = 6.2 x 10^-8

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.

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