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The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.

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M. Sc in physics , teaching since 2 years at home

disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+) CH3COOH ⇔CH3COO- + H+ given: disassociation or ionization constant , K=1.74 ×10 -5 concentration, C =0.05 M So,degree of disassociation or concentration of acetate ions (α) is K=α2C ⇒α...
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disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+)

CH3COOH ⇔CH3COO+ H+

given: disassociation or ionization constant , K=1.74 ×10 -5

concentration, C =0.05 M

So,degree of disassociation or concentration of acetate ions (α) is

K=α2C ⇒α =√(K/C)

α =√(1.74 ×10-5/0.05) =√ 0.000348=0.01865

also ,[H+] =αC =0.01865 ×0.05=9.325×10-4

So pH = -log10[H+]= -log10(9.325×10-4)=3.03

Hence ,pH of solution is 3.03

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