Question #4589c

Mar 18, 2015

The solution will be acidic.

You're dealing with dimethylammonium chloride, which will dissociate in aqueous solution into ${\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+}$ and $C {l}^{-}$.

${\left(C {H}_{3}\right)}_{2} N {H}_{2} C l \left(a q\right) \to {\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+} \left(a q\right) + C {l}_{\left(a q\right)}^{-}$

Notice that the cation is the conjugate acid of a weak base, dimethylamine, ${\left(C {H}_{3}\right)}_{2} N H$, and the anion is the conjugate base of a strong acid, $H C l$.

This means that the reaction of interest, after dimethylammonium chloride dissociates, will be

${\left(C {H}_{3}\right)}_{2} N {H}_{2}^{+} \left(a q\right) + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {\left(C {H}_{3}\right)}_{2} N H \left(a q\right) + {H}_{3} {O}_{\left(a q\right)}^{+}$

The reaction increases the concetration of hydronium ions, which in turn will make the solution acidic.

As a conclusion, look at the cations and anions that form your salt. In this case, you were dealing with an acid salt. Here's a link detailing the acid/base properties of salts

http://www2.onu.edu/~s-bates/chem172/ABSalts12.pdf