Chemistry, 12.03.2020 17:28 kidzay

When a small amount of acid is added to a non-buffered solution, there is a large change in pH. Calculate the pH when 26.9 mL of 0.0044 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

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Answer from: comaslaz21

The correct answer is pH= 2.93

Explanation:

HCl is a strong acid. That means it dissociates completely in water as follows:

HCl (aq) → H⁺(aq) + Cl⁻(aq)

As the dissociation is complete, the amount of H⁺ generated is equal to the concentration of HCl. So, we multiply the volume of HCl added (26.9 mL) by the concentration of HCl (0.0044 M) and then we divide into the volume of water (100.0 mL). Notice that the factor 26.9 mL/100 mL is the dilution factor:

[HCl]= 26.9 mL/100 mL x 0.0044 M = 1.18 x 10⁻³ M

We know that pH= -log [H⁺] and that in this case [H⁺]= [HCl] (because HCl is a strong acid), so we directly calculate the pH from [HCl] as follows:

pH= -log [H⁺]= -log [HCl] = -log (1.18 x 10⁻³ M) = 2.92

Notice that 2.92 is a low value of pH. Hence, when we add a strong acid to water (without a buffer solution), the pH decreases abruptly.

Answer from: Quest

Answer:

5.6⋅10−^19 j

do you need the explaination?

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ 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: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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