chemistry

Calculate the pH of a solution (to 2 decimal places) which is 0.106 M in phenol, Ka = 1.0 x 10-10.

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  1. Let's let HPh equal phenol with H representing the hydrogen that ionizes and Ph representing the remainder of phenol.
    HPh ==> H^+ + Ph^-

    Write the Ka expression.
    Ka = 1 x 10^-10 = [(H^+)(Ph^-)]/(HPh)

    Before the ionization (HPh) = 0.106 M.
    Before the ionization (H^+)=(Ph^-)=0
    After ionization, (H^+)= x
    (Ph^-) = x
    (HPh) = 0.106 - x

    Plug these variables into the Ka expression and solve for x = (H^+).
    Then convert (H^+) to pH with pH = - log(H^+). Post your work if you get stuck.

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  2. pH= -log(1.0x10^-10) + log x/0.106-x.
    how??

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