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25 ml of 0.1 ...

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25 ml of 0.1 M acetic acid is titrated with 0.1NaoH solution. The pH of the solution atequivalence point will be (log 5 0.7CH3COOH = 4.76)

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Q. When 10 mL of 0.1 M acetic acid

(p K_{a} = 5.0)

is titrated against 10 mL of 0.1 M ammonia solution

(p K_{b} = 5.0)

the equivalence point occurs at pH of:

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) is titrated against

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ammonia solution

(p K_{b} = 5.0)

, the equivalence point occurs at pH:

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concentration at the equivalent point when a solution of

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K_{a}

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(b) Variation of degree of dissociation

α

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(c) M acetic acid solution is titrated against M NaOH solution. the difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.

Q. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a 0.1 M NaOH solution (given that

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