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If the Ka of a monoprotic weak acid is 4.8×10−6, what is the pH of a 0.23 M solution of this acid?

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  • 1 month ago
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    HA = H + A-

    ice table needed 

    I ... 0.23 ...0 ..0

    C.....-x......+x..+x

    E.....0.23-x..+x..+x

    Kc= x^2/ .0.23-x = 4.8×10−6

    Kc is very small so very little of the acid dissociates so    .0.23-x ~ 0.23

    x^2 = 0.23 *  4.8×10−6 =  1.10E-06

    x= 1.05 *10^-3 = [H+] from the ice table 

    pH = -log( 1.05 *10^-3) = 2.98  <<<<<< your answer 

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