The calorimeter contains 330 g of water at 31°C. A sample of the substance with a mass of 115 grams is placed into the calorimeter. Once the substance has completely dissolved, the temperature of the water inside the calorimeter is 27.4°C. Recall that the specific heat of water is 4.18 J/g-°C. What is the enthalpy of fusion (Hf) of this sample in joules per gram?

The chemical bubbles

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.

Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m