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pH

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:

10 m L

0.1 M

p K_{a} = 5.0

) is titrated against

10 m L

0.1 M

ammonia solution

(p K_{b} = 5.0)

, the equivalence point occurs at pH:

{O H}^{-}

concentration at the equivalent point when a solution of

0.1 M

acetic acid is titrated with a solution of

0.1 M

N a O H

K_{a}

= 1.9 \times 10^{- 5}

Q. (a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCI solution the variation of pH of solution with volume of HCI added will be :
(b) Variation of degree of dissociation

α

with concentration for a weak electrolyte at a particular temperature is best represnted by :
(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

p K_{a}

of formic acid = 3.74, log 5 = 0.7)

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