Calculate the ph of a solution that is 0.26 m in hf and 0.12 m in naf.

When a mixture consists of a weak acid, such as HF, and its corresponding salt, NaF, this is an acid buffer. So, we can use the equation as written below:

pH = pKa + log(salt/acid)

For HF, the pKa = 3.17

Thus,
pH = 3.17 + log(0.12/0.26)
pH = 2.83

C. chemical changeshope i : )

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.