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Allisa [31]
3 years ago
10

Predict how each molecule or ion would act, in the Brønsted-Lowry sense, in aqueous solution by writing "acid," "base," "both,"

or "neither" on the line provided.

Chemistry
2 answers:
Pavel [41]3 years ago
7 0

The question is incomplete. The molecules/ ions are:

Bicarbonate ion

N-methyl-3-phenyl-3-[4-(trifluoromethyl)phenoxy]propan-1-amine (Prozac)

Para aminobenzoic acid (PABA)

Trinitrotoluene

N-methylpyridinium ion

Answer:

Bicarbonate ion- base

N-methyl-3-phenyl-3-[4-(trifluoromethyl)phenoxy]propan-1-amine (Prozac)-acid

Para aminobenzoic acid (PABA)- both

Trinitrotoluene- acid

N-methylpyridinium ion- neither

Explanation:

In the Brønsted-Lowry sense, an acid is any substance that donates hydrogen ion while a base is any substance that accepts hydrogen ions. Looking at the compounds given above, acidity depends on the molecule's ability to accept or donate protons hence the classification given in the answer.

Fed [463]3 years ago
6 0

Answer:

Both

Explanation:

produce OH− (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton, as we demonstrated in Chapter 4 "Reactions in Aqueous Solution". Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, discussed in Chapter 8 "Ionic versus Covalent Bonding", focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.

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Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for wh
Elanso [62]

Answer: The pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

Explanation:

Given: Initial concentration of Cu(H_{2}O)^{2+}_{6} = 0.10 M

K_{a} = 1.0 \times 10^{-8}

Let us assume that amount of Cu(H_{2}O)^{2+}_{6} dissociates is x. So, ICE table for dissociation of  Cu(H_{2}O)^{2+}_{6}  is as follows.

                               Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of K_{a} is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for K_{a} value is as follows.

K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}

Hence, [H_{3}O^{+}] = 3.2 \times 10^{-5}

Formula to calculate pH is as follows.

pH = -log [H^{+}]

Substitute the values into above formula as follows.

pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49

Thus, we can conclude that the pH of 0.10 M Cu(NO_{3})_{2}(aq) is 4.49.

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