Question

A question asks to find out the pH at the half-neutrilization point. Can someone please explain what this is?

I think it should be that half the moles of the base NH3(being titrated with a strong acid, HI) have been neutrilized, but if this is so, I'm still stuck on how to approach the problem.

We only know the concentration+ volume of NH3 and the conc. of HI.

Answer 1

For a weak acid, HA, being titrated with a strong base, such as NaOH, we have the equation,
HA + NaOH ==> NaA + H2O.

Now what is the Ka for the acid?
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Let's solve for (H^+). That is
(H^+)=Ka*[(HA)/(A^-)]
Now let's turn our attention to the equation. Suppose we start with 2 moles HA. So when we get half way to the equivalence ponit (the equivalence point is when we have added 2 moles NaOH), we will have 1 mole HA left. Right? And how much NaA will have formed. Of course that's 1 mole has been formed because that's how much HA has been titrated. Now look at the equation. Plug in 1 for HA (that's whats left) and 1 for A^- (that's how much of the salt that's formed) and (H^+) = Ka. Neat, huh? It always works that way. When you are half-way to the equivalence point of a weak acid, (H^+) = Ka. When you are half-way to the equivalence point of a weak base, (OH^-) = Kb.