Chemistry, 04.04.2020 04:40 genyjoannerubiera

The acid form of an indicator is yellow and its anion is blue. The Ka of this indicator is 10-5. What will be the approximate pH range over which this indicator changes color?

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Answer from: Quest

the correct answer would be d. hope this !

Answer from: Quest

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

Answer from: Quest

the formula for the conversion of temperature from degree celsius to kelvin is:

$k = ^{o}c + 273.15$

the sample of water at $38^{o}c$ is converted to kelvin temperature by using the above formula as:

$k = 38^{o}c + 273.15 = 311.15 k$

on comparing the two samples of water at $38^{o}c$ or we can say that at $311.15 k$ and at $295 k$ respectively. the difference between the temperatures of two samples of water is:

$311.15 k - 295 k = 16.15 k$

hence, one can say that the sample of water at $38^{o}c$ is greater than the sample of water at $295 k$ by $16.15 k$ .

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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The acid form of an indicator is yellow and its anion is blue. The Ka of this indicator is 10-5. Wha...

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