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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 (a) 0.01M (b) 0.1M in HCl?
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The correct answer is:
α = 1.90, (a) α = 1.82 * 10-5(b) α = 1.82 * 10-4
Correct Option is : (A)
α = 1.90, (a) α = 1.82 * 10-5(b) α = 1.82 * 10-4
α = 1.90, (a) α = 1.82 * 10-5(b) α = 1.82 * 10-4
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