This question has multiple parts. Work all the parts to get the most points.

Several acids are listed here with their respective equilibrium constants:

[tex]\[
\begin{array}{l}
C_6H_5OH (aq) + H_2O (\ell) \rightleftarrows H_3O^+ (aq) + C_6H_5O^- (aq) \\
K_s = 1.3 \times 10^{-10} \\
CH_3CO_2H (aq) + H_2O (\ell) \rightleftarrows H_3O^+ (aq) + CH_3CO_2^- (aq) \\
K_s = 1.8 \times 10^{-5} \\
HCO_3^- (aq) + H_2O (\ell) \rightleftarrows H_3O^+ (aq) + CO_3^{2-} (aq) \\
K_s = 4.8 \times 10^{-11}
\end{array}
\][/tex]

a. Rank these acids according to acid strength, from weakest to strongest.

From weakest to strongest:

[tex]\[ \text{Drag and drop your selection from the following list to complete the answer:} \][/tex]

[tex]\[ CH_3CO_2H \][/tex]

[tex]\[ C_6H_5OH \][/tex]

[tex]\[ HCO_3^- \][/tex]



Answer :

To rank the acids according to their strength, we need to consider their equilibrium constant (K_s) values. The acid strength is directly related to its equilibrium constant value: the larger the K_s value, the stronger the acid, because it dissociates more in water.

Here are the given equilibrium constants:
- [tex]\( C_6H_5OH \)[/tex]: [tex]\( K_s = 1.3 \times 10^{-10} \)[/tex]
- [tex]\( CH_3CO_2H \)[/tex]: [tex]\( K_s = 1.8 \times 10^{-5} \)[/tex]
- [tex]\( HCO_3^{-} \)[/tex]: [tex]\( K_s = 4.8 \times 10^{-11} \)[/tex]

Now let's compare these values from smallest to largest:
- [tex]\( 4.8 \times 10^{-11} \)[/tex]
- [tex]\( 1.3 \times 10^{-10} \)[/tex]
- [tex]\( 1.8 \times 10^{-5} \)[/tex]

Based on these K_s values, we can rank the acids according to their acid strength:

1. [tex]\( HCO_3^{-} \)[/tex]: [tex]\( K_s = 4.8 \times 10^{-11} \)[/tex] (weakest)
2. [tex]\( C_6H_5OH \)[/tex]: [tex]\( K_s = 1.3 \times 10^{-10} \)[/tex]
3. [tex]\( CH_3CO_2H \)[/tex]: [tex]\( K_s = 1.8 \times 10^{-5} \)[/tex] (strongest)

Therefore, the ranking from weakest to strongest acid is:
[tex]\( HCO_3^{-} \)[/tex], [tex]\( C_6H_5OH \)[/tex], [tex]\( CH_3CO_2H \)[/tex]