Question

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

Answer 1

Answer:

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.

Best regards!