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When 10ml of 0.1 M acetic acid (pK(a)=5....

When `10ml` of `0.1 M` acetic acid `(pK_(a)=5.0)` is titrated against `10 ml` of `0.1 M` ammonia solution `(pK_(b)=5.0)`, the equivalence point occurs at `pH`

A

5

B

6

C

7

D

9

Text Solution

Verified by Experts

The correct Answer is:
C
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What will be the pH of 0.1 M ammonium acetate solution ? pK_a=pK_b=4.74

50ml of a solution which is 0.05M in the acid HA (pK_(a) = 3.80) and 0.1M in HB(pK_(a) = 8.20) is titrated with 0.2 M-NaOH solution. Calculate the pH of solution at the first equivalence point. Given your answer after multiplying with 100.

Knowledge Check

  • When 10ml of 0.1 M acetic acid (pK_(a)=50) is titrated against 10 ml of 0.1 M ammonia solution (pK_(b)=5.0) , the equivalence point occurs at pH

    A
    `5.0`
    B
    `6.0`
    C
    `7.0`
    D
    `9.0`

  • When 10 ml of 0.1 M acetic acid (pK_(a) = 5.0) in titrated against 10 ml of 0.1 ml of 0.1 M ammonia solution (pK_(b) = 5.0) , the equivalence point occurs at pH

    A
    `5.0`
    B
    `6.0`
    C
    `7.0`
    D
    `9.0`

  • When 10 ml of 0.1 M acitec acid (pk_(a)=5.0) is titrated against 10 ml of 0.1M ammonia solution (pk_(b)=5.0) ,the equivalence point occurs at pH

    A
    `5.0`
    B
    `6.0`
    C
    `7.0`
    D
    `9.0`

    • Similar Questions

    Explore conceptually related problems

    Calculate the pH at equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M NaOH solution . K_(a) for acetic acid = 1.9xx10^(-5)

    100mL of 0.2M benzoic acid (pk_(a)=4.2) is titrated using 0.2 M NaOH pH after 50mL and 100mL of NaOH have been added are :

    20.0 L of 0.2 M weak acid (pK_(a)=5.0) is titrated against 0.2 M strong base. What is the pH at the equivalence point ?

    pH of when 50mL of 0.10 M ammonia solution is treated with 50 mL of 0.05 M HCI solution :- (pK_(b) "of ammonia"=4.74)

    pH when 100 mL of 0.1 M H_(3)PO_(4) is titrated with 150 mL 0.1 m NaOH solution will be :

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