# Chemistry

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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

• Chemistry -

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.

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