25 mL of a 0.10 M solution of the weak acid hydrofluoric acid, HF, (Ka = 7.2 x 10-4) is titrated with a 0.10 M solution of the weak base ammonia, NH3 (Kb = 1.8 x 10-5). Will the pH at the equivalence point be less than 7, equal to 7, or greater than 7? Explain.

I’m guessing it’s going to be less than 7 because the greater amount of H+ produced from HF since Ka is greater? Or is that wrong??
Or maybe it would be equal to 7 since that's the place where the two are equal??

Wrong^

The eq. point will be below 7 because the Ka is greater than the Kb --> [H+] is greater than [OH-].

The pH at the equivalence point of a titration can be determined by comparing the acid dissociation constant (Ka) of the weak acid and the base dissociation constant (Kb) of the weak base.

In this case, hydrofluoric acid (HF) is a weak acid, and ammonia (NH3) is a weak base. The Ka for HF is 7.2 x 10^-4, and the Kb for NH3 is 1.8 x 10^-5.

At the equivalence point of the titration, the moles of acid will be equal to the moles of base. A neutral salt is formed from the reaction between the weak acid and weak base.

In this titration, HF is the limiting reactant since it is present in a smaller amount (25 mL of 0.10 M). Therefore, the moles of HF will completely react with the moles of NH3.

Since HF is a weak acid, it will partially dissociate to form H+ ions and F- ions. At the equivalence point, all the HF will be neutralized, and the F- ions will combine with the NH4+ ions formed from the reaction between NH3 and H+ ions.

The resulting salt, NH4F, is the conjugate acid of a weak base (NH3) and will hydrolyze in water.

The hydrolysis of NH4F can be represented as follows:

NH4F + H2O ⇌ NH4OH + HF

The ammonium ion (NH4+) will act as an acid (donating an H+ ion), and the F- ion will act as a base (accepting an H+ ion).

Since the Kb of NH3 is smaller than the Ka of HF, the hydrolysis reaction will favor the formation of the weaker acid-conjugate base pair (NH4OH and HF) instead of the stronger base-conjugate acid pair (NH3 and F-).

As a result, more H+ ions will be present in the solution, making it acidic. Therefore, the pH at the equivalence point of this titration will be less than 7.

To determine the pH at the equivalence point of a weak acid-strong base titration, you need to consider the ionization constants of the weak acid (Ka) and the weak base (Kb). In this case, we have a weak acid, hydrofluoric acid (HF), being titrated with a weak base, ammonia (NH3).

At the equivalence point, the moles of acid and base are equal, meaning they completely neutralize each other. In this case, the moles of hydrofluoric acid (HF) and ammonia (NH3) will react to form their corresponding salt, ammonium fluoride (NH4F).

In the first step of the reaction, NH3 accepts a proton from HF, forming NH4+. This is the conjugate acid of NH3, and it is acidic in nature. So, we have NH4+ as the cation.

In the second step, the HF donates a proton to OH- formed by the reaction of NH3 with water. This forms F- as the anion.

NH4F is a salt, so it dissociates completely in water, providing NH4+ and F- ions. Both ions are the conjugate acid and conjugate base of a weak acid and a weak base, respectively.

Now, if we compare the ionization constants, we see that the Ka of HF is greater than the Kb of NH3 (7.2 x 10^-4 > 1.8 x 10^-5). This means that the acid is stronger than the base.

Since the acid is relatively stronger, the resulting solution will be slightly acidic at the equivalence point. Hence, the pH at the equivalence point will be less than 7.

To summarize, the pH at the equivalence point of the titration of HF with NH3 will be less than 7 because the acid (HF) is stronger than the base (NH3), based on the comparison of their ionization constants.

THe pH at equivalence point is always 7