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Standard XII

Chemistry

Lewis Theory

What will be ...

Question

What will be the degree of ionisation of $0.05 M$ acetic acid if its $p K_{a}$ value is $4.74$ ?

0.019

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1.9

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3.0

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4.74

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Solution

The correct option is B $1.9$ %
$p K_{a} = - log K_{a} = 4.74$
$∴ K_{a} = a n t i log (4.74) 1.82 \times 10^{- 5}$
From $K_{a} = \frac{C α^{2}}{(1 - α)} = C α^{2}$ $(1 - α = 1)$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.82 \times 10^{- 5}}{0.05}} = 0.019 = 1.9$ %

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Similar questions

Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
0.01 M

0.1 M in HCl?

Q. Calculate the degree of ionization of

0.05 M

acetic acid if its

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4.74

. How is the degree of dissociation, affected when its solution is (a)

0.01 M

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4.74

. How is the degree of dissociation affected when its solution is also (a)

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and (b)

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in hydrochloric acid?

Q. If

K_{a} = 1.8 \times 10^{- 5}

for acetic acid, its degree of dissociation for 0.05 M solution would be:

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. What should be the ratio of concentration of acetic acid and acetate ions to have a solution with

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What will be the degree of ionisation of 0.05M acetic acid if its pKa value is 4.74?

The correct option is B 1.9%pKa=−logKa=4.74∴Ka=antilog(4.74)1.82×10−5From Ka=Cα2(1−α)=Cα2 (1−α=1)α=√KaC=√1.82×10−50.05=0.019=1.9%

What will be the degree of ionisation of $0.05 M$ acetic acid if its $p K_{a}$ value is $4.74$ ?

The correct option is B $1.9$ %
$p K_{a} = - log K_{a} = 4.74$
$∴ K_{a} = a n t i log (4.74) 1.82 \times 10^{- 5}$
From $K_{a} = \frac{C α^{2}}{(1 - α)} = C α^{2}$ $(1 - α = 1)$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.82 \times 10^{- 5}}{0.05}} = 0.019 = 1.9$ %