pH in Titration?

Question

What are the steps in getting the answer to these kinds of question? How about if I only had the Molarity and the Ka of the acid?

What is the pH of a solution in which 20.0 mL of 0.10 M NaOH are added to 25.0 mL of 0.10 M HCl?
1) The answer is: 1.95.

Answer 1

To solve this you need two things: a. the quantity of excess acid and b. the volume of the resulting solution.

When you have these, you then calulate the solution's molarity.

Finally you use the pH equation to solve for pH.

You know that the solution's pH will be below 7 as there is excess acid present.


(a) At the point when the 20. ml of NaOH were added, you still had excess moles of HCl equal to about 5 x 10-4 moles (that is .005 L x [.10]).

(b) The volume containing this acid was 20 ml + 25 ml =45 ml or .045 L.

Recall that pH = -log[H+]. So we next find [H+].

The [ H+] of this solution is found by dividing the moles of excess acid by the volume.

[H+] = [(5 x 10-4)/[.045] = [1.11 x 10-2]

Then we solve the pH equation for pH.

pH = -log[1.11 x -2] = 1.95.

The Ka problems are more difficult.

Hope this helps and I may have time to help you latter with the Ka's.

Answer 2

Work out how much of the acid has been neutralized. The rest is responsible for the H+ in pH = - log[H+]

HCl is a strong acid, so it is fully dissociated. But of course you need the amount of base added or you wouldn't know how much HCl is left.

If you had an incompletely neutralised weak acid, you would have a more complex situation - a buffer, in fact.

Answer 3

You don't need or want the Ka of the acid. It is a strong acid. The base is a strong base. That makes calculations easier than they otherwise would be.
Excess H ions come from 5ml of 0.1 M HCl. Total volume is 45 ml.
Phase 1:
1 litre of 1MHcl gives 1 mole of H ions
1 litre of 0.1MHcl gives 0.1 moles of H ions
1 ml of 0.1MHcl gives 0.1/1000 moles of H ions
5 ml of 0.1MHcl gives 0.5/1000 moles of H ions = 5X 10^(-4) moles of H ions
Phase II
Concentration of H ions per ml = [5X10^(-4)]/45
Concentration of H ions per litre = [5,000X10^(-4)]/45
= 0.5/45
=0.1/9
= 0.0111111111
Phase III
pH = -log(base10)(H ion conc)
= -log10(0.011111) = 1.95