pH at equivalence point is 9, volume of added base at equivalence point is 30 ml
ID: 905877 • Letter: P
Question
pH at equivalence point is 9, volume of added base at equivalence point is 30 ml.
3. At what volume of added base is the pH calculated by working an equilibrium problem based on the initial concentration and Ka of the weak acid?
4. At what volume of added base does pH=pKa?
Express your answer using two significant figures.
5. At what volume of added base is the pH calculated by working an equilibrium problem based on the concentration and Kb of the conjugate base?
14 12 10 4 0 20 40 60 80 Volume of base added (mL)Explanation / Answer
Solution :-
pH at equivalence point is 9, volume of added base at equivalence point is 30 ml.
3. At what volume of added base is the pH calculated by working an equilibrium problem based on the initial concentration and Ka of the weak acid?
Solution :- When the Base is added to the weak acid then solution forms the buffer solution there fore from the volume of base 0 to before 30 ml we can use the ka of the acid to calculate pH of the solution.
4. At what volume of added base does pH=pKa?
Solution :- When the titration reaches the half equivalence point then moles of the acid and its conjugate base are same therefore the ratio of the base / acid concentrsation is 1
we use the Henderson equation to calculate the pH of the buffer
pH = pka + log ([base]/[acid])
pH= pka + log 1
pH= pka + 0
pH= pka
So at the half equivalence point the pH of the solution is equal to pka
in the titration 30 ml base was needed to reach the equivalence point so 30/2 = 15 ml base needed to reach the half equivalence point therefore at the volume of 15 ml of the base added the pH = pka
so the volume = 15 ml
5. At what volume of added base is the pH calculated by working an equilibrium problem based on the concentration and Kb of the conjugate base?
Solution :-
When the titration reaches to the equivalence point then all the acid is converted to conjugate base
therefore to find the pH of the solutuion we need to use the kb of the conjugate base
therefore the volume is 30 ml of base at which we need to use the kb of the conjugate base needed to calculate the pH of the solution.
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