Question

1. Without doing any math, which solution has a lower pH: (a) 0.05 M acetic acid, or (b) a buffer prepared from 0.05 acetic acid and 0.05 M sodium acetate. Explain your answer. 2. Without doing any math, which solution has a lower pH: (a) 0.05 M acetic acid that has been titrated to its endpoint using NaOH, or (b) a buffer prepared from 0.05 acetic acid and 0.05 M sodium acetate. Explain your answer. 3. What pH did you get for Question 5 in the titration activity? Explain why this answer will always hold true for a strong acid and strong base titration. 4. For the titration of a weak acid, at the half-equivalence point, what ions are present in the solution? 5. What did you get for the pH at the half-equivalence point? 6. What is the pka of the acid being titrated? 7. What sort of solution do you have at the half-equivalence point for the titration of a weak acid?

Answer 1

b/w 0.05m acetic acid & and butter prepared from 0.05m sonham acetatie alid and 0.05m Sodium acetate, 0.05m aletic acid have lower pf Since acetic acid is a weak acid in presence of sodium acetate dissociation of sodium acetate will decrease so [ht] in case of buffer solution ( [tt in acetic acid Solution acid so lower So As because Ctt in echtz acetic solution is greatest it will have pld that is more acidie. When 0.05m acetic acid has been titrated to its endpoint using NaOH, pH of the solution reach canivalence point.. But in buffer solution of acetic acid & Sodium acetate have higher concentration of Ht so butter solution has a lower рн 6 . For the titration of a weak oid at the half eauivalence point, Ht and Conjugate base, acid, will be present in the solution of weak

from henderson equation pH = pka & log [salt] at half equivalence point so pH =pka + logi [salt] =[Acid] => pH = pka.
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