7.53. 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?
7.53. 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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1 Answer
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pKa? =? logKa= 4.74
Ka? = 10? pKa =10?4.74 = 1.8*10?5
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation,x= (Ka / C)1/2? = (1.8*10?5 / 0.05)1/2 ? = 0.019
(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentration from ionization of acetic acid. The hydrogen ion concentration from ionization of HCl is 0.01 M. The total hydrogen ion concentration[H+] = 0.01 + x
The acetate ion concentration is equal to the hydrogen ion concentration from ionization of acetic acid. This is also equal to the concentration of acetic acid that has dissociated.
[CH3?...more
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