how do you find the pH of a titration solution at its halfway point to the equivalence, and how would it affect solutions dealing with strong acids and strong bases, this is an example just to see it quantitatively:
100.0 mL of 0.100 M HF (Ka = 7.2 x 10-4) by 0.100 M NaOH
The pH at the halfway point of a weak acid/srong base or weak base/strong acid is pH = pKa.
Look at the Ka expression for a weak acid.
Ka = (H^+)(A^-)/(HA).
Now solve this for H^+.
(H^+) = Ka*[(HA)/(A^-)].
When you are halfway there, the (HA) = (A^-); therefore, Ka = (H^+). If you take the -log of both sides we get
-logKa = -log(H^+).
By definition the left side is pKa and the right side is pH.
For the titration of a strong acid/strong base, there is no Ka and this rules doesn't apply.
For your example,
millimoles HF = 100 mL x 0.1 = 10 mmols initially.
Halfway there is 50 mL NaOH x 0.1 = 5 mmols.
So you have 5 mmols of the salt formed (NaF) and you have 10.0-5.0 = 5.0 millimols HF remaining. Plug those into the equation I had above and you see one is in the numerator and one in the denominator so they cancel and (H^+) = Ka.
To find the pH of a titration solution at its halfway point to the equivalence, you need to determine the number of moles of the acid and the base that have reacted. The halfway point to the equivalence occurs when exactly half of the moles of the acid have reacted with the same number of moles of the base.
Here's how you can calculate the pH of the halfway point in the given example:
1. Determine the number of moles of HF initially present:
Moles = volume (L) x concentration (mol/L)
Moles of HF = 0.100 L x 0.100 mol/L = 0.010 mol
2. Calculate the number of moles of NaOH required to react with half of the moles of HF:
Moles of NaOH required = 0.010 mol / 2 = 0.005 mol
3. Convert the moles of NaOH to volume using its concentration:
Volume of NaOH required = moles / concentration = 0.005 mol / 0.100 mol/L = 0.050 L or 50.0 mL
4. Calculate the remaining volume of the titration solution:
Remaining volume = initial volume - volume of NaOH added
Remaining volume = 100.0 mL - 50.0 mL = 50.0 mL
5. Determine the concentration of HF after the addition of NaOH:
Concentration of HF = moles / volume = 0.005 mol / 0.050 L = 0.100 mol/L
6. Use the given Ka value (acid dissociation constant) of HF to calculate the pH of the solution:
Ka = [H+][F-] / [HF]
Since the concentration of HF is equal to the concentration of H+, we have:
Ka = [H+]^2 / [HF] (assuming all HF dissociates)
[H+]^2 = Ka * [HF] = (7.2 x 10^-4) * (0.100 mol/L) = 7.2 x 10^-5
[H+] = √(7.2 x 10^-5) = 8.49 x 10^-3 mol/L
7. Calculate the pH using the concentration of H+:
pH = -log[H+]
pH = -log(8.49 x 10^-3) = 2.07
At the halfway point in this titration, the pH of the solution is 2.07.
Now, let's discuss how this relates to solutions dealing with strong acids and strong bases. In the given example, HF is a weak acid, and NaOH is a strong base. At the halfway point of the titration, half of the moles of HF have reacted with equal moles of NaOH. This means that the concentration of HF has reduced to half, resulting in a lower concentration of H+ ions, and therefore a higher pH compared to the initial pH of the HF solution.
In the case of strong acids and strong bases, their titration curves have different shapes. Strong acids fully dissociate in water to produce a high concentration of H+ ions, while strong bases completely dissociate to produce a high concentration of OH- ions. When titrated together, the resulting solutions will have a neutral pH at the equivalence point (the point where the moles of acid and base are stoichiometrically equal), as the H+ ions from the acid react with the OH- ions from the base to form water. Before the equivalence point, the pH will be determined mainly by the acid, usually close to its initial pH, and after the equivalence point, the pH will be determined mainly by the excess OH- ions, resulting in a higher pH.
In summary, the pH of the halfway point in a titration solution is calculated by determining the remaining concentration of the acid and using its acid dissociation constant. For strong acids and strong bases, the pH at the halfway point will be closer to neutral compared to the initial pH of the acid or base solution.