How do I calculate the unknown concentration during titration?
A vinegar cleaner of unknown concentration is titrated with sodium hydroxide solution (c=0.5 mol/l). The solution reaches pH = 7 upon addition of 40 ml of NaOH solution.
a) Calculate the concentration c of the vinegar cleaner.
b) Determine the pH value of the vinegar cleaner
Can someone please help me and explain how to calculate the concentration? I'm wondering if the volume of the vinegar is actually missing.
(2 votes)
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What is pH 7? The EQUIVAL center. This means that the amount of material of the acetic acid and of NaOH is equivalent to aka.
n(ethyl)=n(NaOH)
But what quantity of substance has NaOH now?
c=n/V , that can be converted to cxV=n. Now you’ve got n. c to be determined, but what about V?
The volume of vinegar is missing, yes. But that’s not a problem.
c is always mol per liter. Per litre! 😀
At the b) you have to look after the pKs. Depending on whether the whole is strong or weak, you must choose your formula for the pH.
Okay, I think I’ve done something wrong._.
c=n/V
0.5 mol/l=n/0.04 l, n would then be 0.02 mol, which are now in the solution.
Means that n(ethyl) must also be 0.02. But otherwise the remaining answer would have to fit.
Thank you very much. Now I have as n = 0.02. So then the concentration of vinegar is 0.02 mol/l?
No one can help because nobody knows what volume was titrated by the vinegar cleaner. Only the amount n of acetic acid can be determined via the equivalence criterion which, however, is not at the indicated pH of 7.0. The weak acetic acid is usually titrated against phenolphthalein, the envelope of which is about 9 in the basic pH range. A part of unreacted acetic acid is present at the neutral point of pH 7.0, which should first be calculated.
The pH of the acetic acid cleaner is calculated with the approximation formula for weak acids.
pH = 1/2 * (pKs -lg(c0))
pKs = 4,75 and c0 = unknown because no volume of the acetic acid cleaner is specified
it was a mistake from the chemistry teacher. He forgot the volume. But thank you for your answer!