Titration question 2 equivalence points. Am i right?

AI Thread Summary
The discussion centers on calculating the molar mass of a weak acid from a titration problem, with a sample weighing 0.149 g requiring 16.80 mL of 0.110 M NaOH to reach the equivalence point, resulting in a calculated molar mass of 86.63 g/mol. Additionally, the pH of the weak acid is given as 4.34 after adding 8.40 mL of NaOH, leading to a query about the acid's dissociation constant (Ka). The participant recalls that at the half equivalence point, pH equals pKa, and calculates pKa as 10^-4.38, yielding a Ka of approximately 4.17 x 10^-5. The response affirms the correctness of the concept but suggests verifying the numerical values independently.
A_Munk3y
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Titration question 2! equivalence points. Am i right?

Homework Statement


this is the originial question.
A 0.149 g sample of a weak acid (HA) requires 16.80 mL of 0.110 M NaOH to completely reach the equivalence point. What is the Molar Mass of the acid?
and i got 86.63 g/mol. I have already made sure that it is right.

the question continues,
If the pH of the weak acid in question 1 (above) is 4,34 after 8.40 ml of the NaOH was added, what is the Ka of the acid?

Homework Equations



i think i remember this. At the half equivalence point, PH=PKa

The Attempt at a Solution



since PH=Pka at half equivalence point, the PH=10-4.38
giving me 4.17x10-5

am i right? :)
 
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Your concept is correct. I did not examine your values, so just trust your concept and you may do your own rechecking of the values.
 
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