calculate the percentage ionization of 0.01M acetic acid in 0.1M HCl Ka of acetic acid is 1.8×10^-

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0.01 M

0.1 M H C l . K_{a}

1.8 \times 10^{- 5}

5.0 m L

0.1 M N a O H

50 m L

0.1 M

p H

(K a = 1.8 \times 10^{- 5})

Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74.

How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?

[H^{+}]

0.01 M

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

0.01 M

(K_{a} = 6.3 \times 10^{- 5})

0.4 M

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

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