How to calculate the pH of a titration? Can someone explain how to find H+ when
ID: 482329 • Letter: H
Question
How to calculate the pH of a titration? Can someone explain how to find H+ when a volume of 100 mL is added and when a volume of 100.01 mL is added? Once I get H+ for these, o find the pH for these, I would just use -log(x) right? Thank you!
l. The following table is for a titration of 100.00 mL of 0.1000M HCl (analyte in flask) with 0. 1000M NaOH (titrant in buret) a. Complete the table below. Show any necessary work. Volume of base added to flask 0 IH (0.1000M) 1.09 10.00 (0.110L 2.28 90.00 3.30 99.00 4.30 99.90 9999 100.00 Hint: consider the dissociation of water 100.01 Hint: calculate excess IOHj 14-pOH 8.7 100.10 101.00 10.7 110.000 11.7Explanation / Answer
When 100 mL base is added the equivalence point is reached, i.e., all HCl converts to NaCl, and no HCl remains in the solution unreacted.
HCl + NaOH --> NaCl + H2O
Thus, H+ ion will be available in the solution only from the small amount of dissociation of water.
H2O <==> H+ + OH-
At room temperature the ionic product of water is 10-14, i.e.,
[H+][OH-] = 10-14
As, in the dissociation of water, [H+]=[OH-]
so, [H+]2 = 10-14
or, [H+] = 10-7
or, pH = -log[H+] = -log10-7
= 7
As there is no HCl left in the solution, so further addition of NaOH only add excess OH- in the solution.
NaOH ---> Na+ + OH-
Extra 0.01 mL NaOH will give OH- ion in the solution
= (0.01 /1000)L * 0.1 M
= 10-6 moles
total volume of the solution = 100 + 100.01 = 200.01 mL
concentration of OH- ion = [(10-6 moles) * 1000] / 200.01 mL
= 5*10-6 M
pOH = -log[OH-]
= -log[5*10-6]
= 5.3
or, pH = 14 - 5.3
= 8.7
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