It requires 19.58 mL of NaOH solution to titrate 0.4307 g of the primary standar

Question

It requires 19.58 mL of NaOH solution to titrate 0.4307 g of the primary standard KHP (MW=204.23g/mol). It requires 16.00 mL of the same NaOH solution to titrate 0.1147 g of an unknown weak acid dissolved in 100.0 mL of water. From the titration curve, the acid appears to be diprotic with pKa1= 2.56 and pKa2=4.37.

a)Calculate the equivalent weight of the unknown weak acid (g/eq).

b)Calculate the molecular weight of the acid described in Question 1 (g/mol).

c)Calculate the pH of the weak acid solution described in Question 1 at the point whereonly 5.00 mL of NaOH have been added.

d)Calculate the pH of the solution desribed in Question 1 at the second equivalence point.

Explanation / Answer

No your answers are not correct. : Hints:
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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