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Identify each species in the following e...

Identify each species in the following equilibrium according to the code:
SA = stronger acid , SB = stronger base , WA = weaker acid , WB = weaker base.
The `pK_a` of `(CH_3)_2NH` is 36, the `pK_a` of `CH_3OH` is 15.2.
`underset(1)(CH_3OH)+underset(2)((CH_3)_(2))NHhArrCH_3-O^(-) +CH_3-underset(H)underset(|)(overset(+)(NH))-CH_3`

A

`{:(1,2),(WA,WB):}`

B

`{:(1,2),(WB,WA):}`

C

`{:(1,2),(SA,SB):}`

D

`{:(1,2),(SB,SA):}`

Text Solution

AI Generated Solution

The correct Answer is:
To identify each species in the given equilibrium according to the specified codes (SA = stronger acid, SB = stronger base, WA = weaker acid, WB = weaker base), we will analyze the provided pKa values and the reaction. ### Step-by-Step Solution: 1. **Identify the Species**: The reaction given is: \[ \text{CH}_3\text{OH} + \text{(CH}_3)_2\text{NH} \rightleftharpoons \text{CH}_3\text{O}^- + \text{CH}_3\text{NH}_3^+ \] 2. **Determine the pKa Values**: - The pKa of \((CH_3)_2NH\) is 36. This indicates that it is a very weak acid because a high pKa value corresponds to a low dissociation constant (Ka). - The pKa of \(CH_3OH\) is 15.2. This indicates that it is a stronger acid compared to \((CH_3)_2NH\) since a lower pKa value corresponds to a higher dissociation constant. 3. **Identify the Acid and Base**: - In the reaction, \(CH_3OH\) donates a proton (H⁺) to \((CH_3)_2NH\), which means \(CH_3OH\) acts as an acid. - The species \((CH_3)_2NH\) accepts the proton and becomes \((CH_3)NH_3^+\), indicating that it acts as a base. 4. **Classify the Acids and Bases**: - Since \(CH_3OH\) has a pKa of 15.2, it is a weak acid (WA). - Since \((CH_3)_2NH\) has a very high pKa of 36, it is a weaker acid (WA) and, when it accepts a proton, it forms \((CH_3)NH_3^+\), which is a conjugate acid. - The conjugate base formed from \(CH_3OH\) is \(CH_3O^-\), which is a stronger base (SB) compared to \((CH_3)_2NH\). 5. **Final Classification**: - \(CH_3OH\) = WA (weaker acid) - \((CH_3)_2NH\) = WB (weaker base) - \(CH_3O^-\) = SB (stronger base) - \(CH_3NH_3^+\) = SA (stronger acid) ### Summary of Identifications: - **CH3OH** = WA (weaker acid) - **(CH3)2NH** = WB (weaker base) - **CH3O^-** = SB (stronger base) - **CH3NH3^+** = SA (stronger acid)

To identify each species in the given equilibrium according to the specified codes (SA = stronger acid, SB = stronger base, WA = weaker acid, WB = weaker base), we will analyze the provided pKa values and the reaction. ### Step-by-Step Solution: 1. **Identify the Species**: The reaction given is: \[ \text{CH}_3\text{OH} + \text{(CH}_3)_2\text{NH} \rightleftharpoons \text{CH}_3\text{O}^- + \text{CH}_3\text{NH}_3^+ \] ...
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