Chemistry, 29.08.2019 16:30 7obadiah

The pk of ch nh,+ equals 10.6; the pk_of (ch3)2nh,* equals 10.7. which is the stronger base, ch nh, or (ch3)2nh²

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Answer from: cabr379

Explanation:

Relation between $pK_{a}$ and $pK_{b}$ is as follows.

$pK_{a} + pK_{b}$ = 14

Hence, putting the given values of $pK_{a}$ in order to calculate the value of $pK_{b}$ as follows.

For $CH_{3}NH^{+}_{3}$ : $pK_{a} + pK_{b}$ = 14

10.6 + $pK_{b}$ = 14

$pK_{b}$ = 14 - 10.6

= 3.4

For $(CH_{3})_{2}NH^{+}_{2}$ : $pK_{a} + pK_{b}$ = 14

10.7 + $pK_{b}$ = 14

$pK_{b}$ = 14 - 10.7

= 3.3

Hence, smaller is the value of $pK_{b}$ of a specie more will be the value of its $K_{b}$ . Therefore, stronger will be the base.

Thus, we can conclude that $(CH_{3})_{2}NH^{+}_{2}$ will be the stronger base.

Answer from: Quest

a. a primary source combines several different accounts of an event. b. a primary source generally benefits from having been written after an event.

explanation:

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 pk of ch nh,+ equals 10.6; the pk_of (ch3)2nh,* equals 10.7. which is the stronger base, ch nh,...

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