Determine the Ka of the unknown weak acid. I am suppose to use the equations: pH

Question

Determine the Ka of the unknown weak acid. I am suppose to use the equations:

pH=-log[h30+]

pka=-log ka

pka=pH

I determined th epka of my unknow acid is 4.2. I determined this as "The average pH at half-equivalence point is equal to the pKa of the unknown weak acid" according to my lab. I took two trials and on trial one I got 4 as the pH for the half equvelence point and on trial two, I got 4.5 has the pH.

My equivelence point on trial one was 8.5 and trial two is 12.5, a huge difference that averaged to 10.5.

I do not understand the equations and how to plug in the numbers. an explanation would be great! Thank you.

Explanation / Answer

that equivalence point pH is to different, specially if it is the same subtance and same volume.

Always apply:

pH = pKa + log(conjugate/Acid)

since in the HALF-equivalence point we have same amount of conjugate= acid

then

pH = pKa + log(1)

pH = pKa

the problem is that you will have a very large volume differenc ebetween 8.5 and 12.5

then

If this was my lab report, I will check the area in which there is buffer formation, i.e. the pH does not changes that much

selet the initial point, the final point and its half point, that will be the pKa = pH

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