0.1 M sodium hydroxide (strong base) is added to 25 ml of 0.1 M of ethanoic acid (weak acid) in a titration. What is the pH at equivalence?

The equivalence point is reached when the titrant added neutralizes the titrand. It is thought that when a solution is neutral it would have a pH of 7. However, the combination of a weak acid and a strong base results in the formation of a salt which when undergoes hydrolysis releases OH- anions into the solution. The contrary is also the case where the combination of a strong acid and a weak base forms a salt which when undergoes hydrolysis releases H+ ions. Thus, to calculate the pH at equilibrium the amount of the salt which is formed must first be calculated before calculating the amount of ions released into the solution altering the pH. To calculate this, the I. C. E. method can be implemented which stands for Initial, Change and Ending concentrations or moles.
For this example, since both the titrant and titrand are at the same concentration they require the same volume to neutralize. Thus, 25 ml of ethanoic acid is reacted with 25 ml of sodium hydroxide. Knowing the neutralization equation, 1 mole of ethanoic acid reacts with 1 mole of sodium hydroxide to form 1 mole of sodium ethanoate and 1 mole of water. Using the equation of the reaction, the concentration of the produced sodium ethanoate is calculated to be 0.05 M.
Following the first step the ICE method is applied to the hydrolysis equation of the salt. As it is unknown the level of completion the variable "x" is utilized in the Change portion of the method. Thus, the end concentration of the sodium ethanoate is calculated to be 0.05 - x , however, the change in the concentration can be concidered negligible and thus the value of 0.05 M is used as the final concentration. The concentration of ethanoic acid and hydroxide is determined as x.
Next, the definition of the dissociation constant of a base is utilized. Where Kb = [BH+][OH-] / [B]. The pKa of ethanoic acid is 4.76 and thus its conjugate base has a pKb of 14-4.76 = 9.24. This pKb can be converted into the Kb through its definition where Kb = 10 ^ - pKb. Thus the Kb is determined to be 5.75 x 10^-10. Thus through calculation the concentration of the hydroxide is determined to be 5.36 x 10^-6 and the pOH is determined to be 5.27 and the pH to be 8.73.