The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.
Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.
The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.
Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.
Of course it isnt Jans fault that the interpretation of the formatting is different on mobile phones. That absolutely isnt his problem. // Stupid !@#$%^&* mobile phones...More
Nice answer. I know what you mean for the equation with Ka,2 but that is a really weird way to write the reaction. I really think you should have written the reactions as two separate equations. $$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ }$$ $$\ce{H3C-NH2 <=>[$K_\mathrm{a2}$] H3C-NH- + H+}$$More
Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.
$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$
The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.
Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.
Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values.
$$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$
The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little practical relevance. Most of the time, the question is not how acidic methyl amine is but how basic. And for that, the $\mathrm{p}K_\mathrm{a1}$ value of the conjugate acid, methyl ammonium, is used. $\ce{H3C-NH3+}$ has a $\mathrm{p}K_\mathrm{a1}$ value of approximately $10$, and it is the value you found online.
Formally correct usage would only name $\mathrm{p}K_\mathrm{a2}$ as the value of methyl amine. Colloquial chemists’ usage mainly uses $\mathrm{p}K_\mathrm{a1}$.
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