A strong base- strong acid titration curve. 0.1 M HCl is being added to 0.1 M NaOH 50 ml. help needed.?

To clarify: You are titrating a strong base (NaOH) with a strong acid (HCl).

The 0.1M NaOH has pH = 13.0 ( NOT pOH. For 0.1M NaOH , pOH = 1.0)

When you start the titration the pH = 13.0

As you add the 0.1M HCl , the pH will drop slowly until all the NaOH has been neutralised . That is when 50mL of 0.1M HCl solution has been added. You then have 100mL of a NaCl solution with pH = 7.00.

The above is a somewhat stylised version of what happens. In practice ,the pH ( at the end point of the titration ) will drop from ≈ 8.5 to ≈ 4.5 with the addition of 1 drop of the 0.1M HCl solution . The line on the graph will be vertical , passing through the end point pH , which is 7.00. There is a practical problem in stopping the pH at exactly 7.00, the end point. Remember that at pH = 4.5, [H+] , and in this case [HCl] = 10^-4.5M = 0.000032M

You should refer to this site for extra clarity on this matter - it deals with indicators, but it describes all types of acid / base titrations very well.

http://www.chemguide.co.uk/physical/acidbaseeqia/i...

Click on what part of the link is visible . K!A has a bad habit of cutting off long entries. But if you click on what is shown you will still go to the site.