Question

A weak acid, HF, is in solution with dissolved sodium fluoride, NaF. If HCl is added, which ion will react with the extra hydrogen ions from the HCl to keep the pH from changing?

Answer 1

Answer:

$F^-$

Explanation:

The weak acid HF is in solution with dissolved NaF; NaF is an ionic compound, so it will dissolve by dissociating its ions; this solution reaction is:

NaF⇒$Na^++F^-$

So, initially there will be these species in solution:

-NaF

-$Na^+$

-$F^-$

If we add HCl, which is a strong acid, it will dissociate completely (that is the characteristic of a strong acid) in ions H+ and Cl-:

HCl⇒$H^++Cl^-$

So, we are finally adding ions $H^+$ and $Cl^-$.

We are asked which ion will react with the $H^+$ released by HCl acid. We should think just about the anions (negative ions) because we require an electron donor to create a bond with H+, so it won't be on solution as H+. We just have two anions in solution: $F^-$ and $Cl^-$.

If $Cl^-$ bonded to $H^+$ it would form HCl, but we saw that HCl is a strong acid which always dissociate completely in aqueous solution; so $Cl^-$ will never bond to $H^+$. Finally, $F^-$ can bond $H^+$ giving HF as a result; it is possible because HF is a weak acid, and HF can be, in fact, present in an aqueous solution; the $H^+$ ions will be changed to HF and the pH (which depends only on $H^+$ concentration) will not change.