Calculate the pH at the equivalence point in titrating 0.110 M solutions of each of the following with 8.0×10−2 MNaOH.

Part A
hydrobromic acid (HBr)

HBr + NaOH ==> NaBr + H2O

Since HBr is a strong acid and NaOH is a strong base, the material at the end is just a salt in water. The pH of the water will be 7 (at least that's the simple answer and it neglects ionic strength and the Debye-Hueckel theory).

To determine the pH at the equivalence point in the titration of hydrobromic acid (HBr) with sodium hydroxide (NaOH), we need to consider the stoichiometry of the reaction between the two compounds and the ionization of the resulting species.

The balanced equation for the reaction between HBr and NaOH is as follows:
HBr + NaOH → H2O + NaBr

At the equivalence point, we have an equal number of moles of HBr and NaOH. This means that all of the HBr has reacted with NaOH, resulting in the formation of water (H2O) and sodium bromide (NaBr). In other words, the HBr is completely neutralized by the NaOH.

Now, we need to determine the concentration of OH- ions that are present in the solution at the equivalence point. Since we have a solution of NaOH with a concentration of 8.0×10^-2 M, we can calculate the concentration of OH- ions using stoichiometry.

In the balanced equation, we can see that for every mole of NaOH we have, one mole of OH- ions is produced. Therefore, the concentration of OH- ions at the equivalence point is also 8.0×10^-2 M.

Now, to calculate the pH at the equivalence point, we need to consider the autoionization of water. Water undergoes self-ionization to produce H+ ions (protons) and OH- ions, according to the equation:

H2O ↔ H+ + OH-

The concentration of OH- ions at the equivalence point is 8.0×10^-2 M. Since the solution is neutral, the concentration of H+ ions will also be 8.0×10^-2 M.

The pH is defined as the negative logarithm (base 10) of the H+ ion concentration. Therefore, we can calculate the pH using the formula:

pH = -log[H+]

Plugging in the concentration of H+ ions, we get:

pH = -log(8.0×10^-2) = 1.10

Therefore, the pH at the equivalence point in the titration of 0.110 M hydrobromic acid (HBr) with 8.0×10^-2 M NaOH is approximately 1.10.