1. Calculate the percent ionized in 6.02x10-6 M phenol--a very dilute solution.

Expert Solution

C6H5OH ---------------> C6H5O^- + H^+

I 6.02*10^-6 0 0

C -x +x +x

E 0.00000602-x +x +x

Ka = [C6H5O^-][H^+]/[C6H5OH]

1.6*10^-10 = x*x/0.00000602-x

1.6*10^-10*(0.00000602-x) = x^2

x = 3.1*10^-8

[H^+] = 3.1*10^-8 M

the percent ionization =H^+ conc *100/inital conc of acid

= 3.1*10^-8 *100/6.02*10^-6 = 0.514% >>>>answer

queen_honey_blossom answered 2 years ago

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