Find the pH for this titration problem.?

Question

A titration was started by taking 20ml of 0.01 molar HCL.Then A SOLUTION OF NaOH WAS GRADUALLY ADDED .By mistake the end point was missed and excess NaOH was added.At the end the solution was 0.0001molar in NaOH.What was the pH of the solution at the beginning,at the end point and at the end of titration?

Answer 1

pH = -log [H+]

At the beginning, [H+] = 0.01 = 10^-2. Log [H+] = -2, so pH = 2

At the end, Kw = [H+][OH-] = 10^-14. So 0.0001M NaOH has 10^-4H OH-. Then [H+] x 10^-4 = 10^-14. [H+] = 10^-10, log [H+] = -10, and pH = 10

Answer 2

You'll need two different formulas for this problem:

pH= -log[H+]
pH+pOH= 14

at the beginning, the concentration is only 1x10^-2 M of HCl ( a strong acid) so you can assume the same concentration for H+. Then the beginning pH would be

pH= -log[1x10^-2] = 2

Now, NaOH is a strong base and HCl is a strong acid. The true end point of any titration with a strong acid and a strong base should be pH=7 or at least very close to it.

As for the end of the titration, all you had remaining was essentially 1x10^-4 M of NaOH left over. Since this is a strong base, you can also assume you have 1x10^-4 M of OH- as well. Calculating the pOH (since you have no H+) gives you

pOH=-log[1x10^-4]=4

and

pH+pOH=14

pH=14-4= 10

so your pH is 10 at the end of the titration.

Hope this is helpful

Answer 3

If the final concentration of the NaOH was 0.0001 mol/dm3 then then use Kw and rearrange to get the Hydrogen ion concentration. Then get pH using pH = -Log H+

Answer pH =10