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March 24, 2017

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calculate the hydronium ion concentration in 0.125M formic acid, HCHO2
Ka= 1.8 x 10-4
can you show me the steps pls

  • chemistry - ,

    ...........HCOOH ==> H^+ + HCOO^-
    initial....0.125M.....0......0
    change.......-x........x......x
    equil.....0.125-x......x......x

    The ICE chart is set up above.
    Ka = 1.8E-4 = (H^+)(HCOO^-)/(HCOOH)
    Substitute from the ICE chart and solve for (H^+).

  • chemistry - ,

    DrBob is the HCOOH---->H^++HCOO^ the HCHO2

  • chemistry - ,

    Yes. HCOOH is formic acid. The H on the right side comes from the -COOH or an organic molecule, called a carboxylic acid, and that can ionize.

  • chemistry - ,

    I'm stuck
    Ka=1.84 x 10-4= x x/ 0.125

    (1.84 x 10-4)[0.125]
    2.3 x 10-5= x = [H30]
    lost which way do I go

  • chemistry - ,

    Ka = (x)(x)/(0.125-x) = 1.84E-4
    We make the problem a little easier to solve if we assume x is small and that 0.125-x = 0.125. Then
    1.84E-4 = x^2/0.125
    1.84E-4*0.125 = x^2
    2.3E-5 = x^2
    x = sqrt(2.3E-5) = 0.00480M = (H3O^+).

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