Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

What is the pHof a weak monoprotic acid in a 0.01482M solution, given its pka=4.20?

2 Answers

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  • Bobby
    Lv 7
    2 months ago

    for  a weak acid [H=] = sqrt ( Ka *c)

    this assumes a very small dissociation of the weak acid 

    this is reasonable given the very small Ka

    so there is no need for ICE tables 

    Ka = 10^-4.2 = 6.310 * 10 ^-5

    [H+] = sqrt( 6.310 *10 ^-5 * 0.01482) =9.67E-04

    pH = log(H+) = -log(9.67E-04 )

    = 3.01

    there is your answer 

  • 2 months ago

    0.01482 =[HA]

     pka=4.20 ; Ka = 6.309 x 10^-5

         HA     ⇔         H+   +    A -

     I   0.01482.          -            -

    C   -y                   +y           +y

    E  0.01482 -y.        y              y

      .: Ka = [H+][A-]/[HA] = y²/(0.01482 -y) = 6.309 x 10^-5

         y= 0.000935 = [H+]

    pH = - log[H+]= - log[0.000935]= 3.03 

    Please rate Thank you.

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