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Salt of Weak Acid and Strong Base

0.1M weak aci...

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0.1M weak acid HA ph=3 is titrated with 0.05M NaOh solution.calculate the ph wen 25% of acid has beem neutralised.

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[H^{+}]

at equivalent point between titration of

0.1 M, 25 m L

H A (K_{a (H A)} = 10^{- 5})

0.05 M N a O H

Q. A weak acid of dissociation constant

10^{- 5}

is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be:

Q. A solution of weak acid was titrated with base

N a O H

. The equivalence point was reached when

36.12 m L

0.1 M N a O H

have been added. Now

18.06 m L 0.1 M H C l

were added to titrated solution, the

p H

was found to be

4.92

K_{a}

0.1 M

solution of a weak acid

H A

N a O H

slowly. The titration curve observed is given below.

k_{a} = 4 \times 10^{- 5}

25^{\circ} C .

The correct observation for the curve will be:

B O H

is titrated with a strong acid

H A

10 m l

H A

p H

9.00

25 m l

p H

8.00

. The volume of the acid required to reach the equivalence point is

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Salt of Weak Acid and Strong Base

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