if 50.0 mL of 1.00 M HC2H3O2 (Ka= 1.8*10^-5) is titrated with 1.00 M NaOH. What
ID: 808237 • Letter: I
Question
if 50.0 mL of 1.00 M HC2H3O2 (Ka= 1.8*10^-5) is titrated with 1.00 M NaOH. What is the pH of the solution after the following volumes of NaOH have been added? a)0.00mL b)25.00 mL c) 49.90 mL d) 50.10 mL [please be as specifc and as dreadful with the details as possible..I am studying for a test and need lots of help, thank you!] if 50.0 mL of 1.00 M HC2H3O2 (Ka= 1.8*10^-5) is titrated with 1.00 M NaOH. What is the pH of the solution after the following volumes of NaOH have been added? a)0.00mL b)25.00 mL c) 49.90 mL d) 50.10 mL [please be as specifc and as dreadful with the details as possible..I am studying for a test and need lots of help, thank you!]Explanation / Answer
CH3COOH is a weak acid - calculate [H+] from the Ka equation for question a)
When you add NaOH to the acid you produce a buffer solution Use the Henderson Hasselbalch equation
At the half equivalence point , pH = pKa . pKa for the acid is 4.77 . The half equivalence point is c) So the pH here will be 4.77
I will leave you to rework the problem . If you have not made any progress , or someone else has not done the work for you , I will come back to this problem tomorrow.
Edit next day:
No progress from you and no other answer: So we will do this in full
Question a) You have a weak acid. In order to calculate pH you first have to determine the [H+] in the solution. You do this using the Ka equation:
Ka = [H+] [CH3COO-] / [CH3COOH]
You know that [H+] = [CH3COO-] so for product we write [H+]
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