The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.460 moles of a monoprotic weak acid (Ka = 3.8 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
To find the pH at the half-equivalence point, we first need to determine the amount of acid that has reacted with the base at that point.
At the half-equivalence point, half of the acid has reacted with the base, which means that the remaining half is still unreacted. Therefore, the amount of acid remaining is 0.460 moles / 2 = 0.230 moles.
Next, we can use the balanced chemical equation between the acid (HA) and the base (NaOH) to determine the number of moles of base required to react with the acid completely. Since the acid is monoprotic, it will react with one mole of base per mole of acid.
The balanced equation is as follows:
HA + NaOH → H2O + NaA
Since the acid is weak, it does not completely dissociate in solution. Therefore, the reaction can be represented as:
HA + OH- → H2O + A-
We can see that one mole of acid reacts with one mole of hydroxide ion (OH-) to form one mole of water (H2O) and one mole of the conjugate base (A-).
Given that the amount of acid remaining is 0.230 moles, we can conclude that 0.230 moles of NaOH are required to neutralize the remaining acid.
To calculate the amount of hydroxide ions (OH-) present in the solution, we need to consider that NaOH is a strong base and dissociates completely in solution. Therefore, the number of moles of OH- ions is equal to the number of moles of NaOH added.
So, the amount of OH- ions present in the solution is also 0.230 moles.
To determine the concentration of hydroxide ions, we need to divide the amount of OH- ions by the volume of the solution at the half-equivalence point. However, since we don't have information about the volume, we cannot calculate the concentration directly.
However, we can still determine the pOH of the solution by using the equation:
pOH = -log[OH-]
Since [OH-] is equal to 0.230 moles (as discussed above), we can substitute this value into the pOH equation:
pOH = -log(0.230)
Calculating this using a calculator, we get:
pOH ≈ 0.638
Finally, to find the pH of the solution at the half-equivalence point, we can use the fact that pH + pOH = 14:
pH + 0.638 = 14
Subtracting 0.638 from both sides of the equation:
pH ≈ 13.362
Therefore, the pH of the solution at the half-equivalence point is approximately 13.362.