Answer:
![F^-](https://img.qammunity.org/2020/formulas/chemistry/middle-school/s4c309pclg4yca3mr855464tlxcu27a3ra.png)
Step-by-step 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^-](https://img.qammunity.org/2020/formulas/chemistry/high-school/np92kz3h2mk03dwk0mt0bj1uqn85uvmlmh.png)
So, initially there will be these species in solution:
-NaF
-
![Na^+](https://img.qammunity.org/2020/formulas/physics/high-school/qckrbxf9jnx6p497x0gerknyu6555t8gj0.png)
-
![F^-](https://img.qammunity.org/2020/formulas/chemistry/middle-school/s4c309pclg4yca3mr855464tlxcu27a3ra.png)
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^-](https://img.qammunity.org/2020/formulas/chemistry/high-school/a3hybcbjvyxw99txi8dzjvz371bibwumlg.png)
So, we are finally adding ions
and
.
We are asked which ion will react with the
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:
and
.
If
bonded to
it would form HCl, but we saw that HCl is a strong acid which always dissociate completely in aqueous solution; so
will never bond to
. Finally,
can bond
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
ions will be changed to HF and the pH (which depends only on
concentration) will not change.