Chemistry, 07.07.2021 01:50 fnaffan

or each conjugate acid-base pair, identify the first species as an acid or base and the second species as the conjugate acid or base of the first. A) CH3COOH and B) CH3COO- A. is B. is C) CH3O- and D) CH3OH C. is D. is

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Answer from: bellamore

A. an acid

B. a base

C. conjugate base

D. conjugate acid

Explanation:

This question appears incomplete because it is starting with a "or".

However, the available question can still be answered.

An acid (according to "Bronsted-Lowry") is a chemical compound/specie that can donate hydrogen ion(s) in a reaction while a base is the compound/specie that can receive hydrogen ion(s). From this definition, it can be deduced that CH₃COOH (ethanoic acid) is an acid (weak acid) because it can donate an hydrogen ion while CH₃COO⁻ is a base because it can receive a hydrogen ion.

Also, a conjugate acid is the compound formed when an acid has donated a hydrogen ion to a base (i.e it is a base with hydrogen ion attached to it) while a conjugate base is the acid that has lost it's hydrogen ion. Thus, the conjugate acid in the question is CH₃OH while the conjugate base in the question is CH₃O⁻.

A. an acid

B. a base

C. conjugate base

D. conjugate acid

Answer from: Quest

30litre of li2o is 1.0002mole approximately 1 mole

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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or each conjugate acid-base pair, identify the first species as an acid or base and the second speci...