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A solution of acetic acid is 1.0% ionized. Determine the molar concentration of acid Ka = 1.8 × 10-5 and also the H+

a
1.8 × 10-1 M and  1.8 × 10-3M
b
0.18 × 10-1 M and  1.8 × 10-4M
c
0.18 × 10-2 M and  1.8 × 10-2M
d
0.18 × 10-3 M and  1.8 × 10-1M

detailed solution

Correct option is A

CH3COOHC - Cα  ⇌  CH3COO-Cα  + H+Cα     where ‘C’ is the concentration of the acid and α is the degree of dissociation.Ka=Cα×CαC(1-α) As α is very small can be neglected in the denominator Ka=Cα2   Ka = 1.8 × 10-5= Cα2∴           1.8 × 10-5= C.0.012or        C=1.8 × 10-5  ×  104               =   1.8 × 10-1MH+ = Cα  or  H+ =1.8 × 10-1 × 0.01         =0.18 × 0.01=0.0018 M  or  1.8 ×  10-3  M

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