College Chemistry

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Calculate the pH of a 0.5 M Formic Acid (HCOOH) soln. ka= 0.00018

  • College Chemistry -

    Ka=[H][COOH]/([HCOOH]

    ka= x^2/(.5-x)

    solve this for x
    x^2+.00018x-.5*.00018=0

    x=(-.00018+-sqrt(.00018^2+.00036))/2

    x=-.00009+- .00948= .00939 check all that math.

    pH=-log (.00939)=2.03

    check the math.

  • College Chemistry -

    The equilibrium is given by

    HCOOH -> H+ + HCOO-

    so Ka=[H+][HCOO-]/[HCOOH]

    if we start with 0.5M HCOOH then at equilibrium there is [H+] = x and [HCOO-]=x and HCOOH=0.5-x

    so Ka = (x)(x)/(0.5-x)=0.00018

    this gives a quadratic to solve, however, we only want the answer to 1 sig fig (.5M is 1 sig fig) so we can say that 0.5-x is approximately equal to 0.5, because x will be small.

    thus x^2/0.5=0.00018

    hence you can find x

    pH is then -log(x)

    which is pH=2, but check the maths.

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