Chemistry, 18.01.2021 14:00, zalyndevola

Propionic acid (CH3CH-COOH) has a K, 1.3 x 10^-5 and phosphoric acid (H3PO2) has a Ka = 7.5 x 10^-3 Choose the conjugate base for each.
A. CH3CH2COO2- for CH3CH2COOH; HPO4 2- for H3PO4
B. CH3CH2CO- for CH3CH2COOH; H2PO3 - for H3PO4
C. CH3CH2COOH2 for CH3CH2COOH; H4PO4 for H3PO4
D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4

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answered: rama64

D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4.

Explanation:

Hello!

In this case, since the ionization of a weak acid produces a conjugate acid, molecule to which the leaving H⁺ is added to, and a conjugate base, everything that remains after the withdrawal of the H⁺, we can note down the ionization of each acid down below:

$CH_3CH_2COOH\rightleftharpoons CH_3CH_2COO^-+H^+\\\\H_3PO_4\rightleftharpoons H^++H_2PO_4^-$

Thus, we notice that the answer is D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4 because the phosphoric acid is most likely to ionize to dihydrogen phosphate ions rather than phosphate, or hydrogen phosphate ions it undergoes a stepwise ionization in which the first step is the predominant one.

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answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.