Suppose we were given a sample of acetic acid, CH_3 COOH (K_a = 1.8 times 10^-5)
ID: 496019 • Letter: S
Question
Suppose we were given a sample of acetic acid, CH_3 COOH (K_a = 1.8 times 10^-5) of unknown concentration. You titrate 25.00 mL of this solution with a 0.55 mol/L NaOH solution. If your initial buret reading was 0.50 mL and your buret reading at your endpoint was 30.60 mL, what was the initial concentration of acetic acid in your 25.00 mL sample? Suppose we plan to repeat the titration from the previous question. Again, we add a 25.00 mL sample of our unknown concentration of acetic acid to a flask and get ready to titrate with NaOH. However, before we start titrating, we add an additional 100 mL of water to the flask containing our sample of acid. Everything else about the titration is the same. Compared to the titration in the previous question.... ... the pH at the start of the titration (before any NaOH is added) will be... ... the concentration of acetate at equivalence will be... ... the volume of NaOH needed to reach equivalence will be... ... the pH at the half-equivalence point will be... higher lower the sameExplanation / Answer
In second experiment we took 25mL of 0.6622M acetic acid solution and added 100mL of water into it.
Q1) The acid is now more diluted, thus the [H+] concentration decreases , hence pH increases. The pH of the ne solution is higher.
Q2)As the concentration of acetic acid is lower, the concentration of acetate at equivalence is is lower.
Q3)The volume of NaOH needed to reach equivalence will be lower as the concentration is lowered .
Q4) The pH at the half equivalence point remains the same., as at this point it behaves as a buffer , whose pH is given by Hendersen equation as
pH = pKa + log [conjugate bae]/[acid]
at half equivalence [conjugate base] = [acid] and thus
pH = pKa irrespective of the actual values of concentrations
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