chemistry

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Calculate the pH of each of the following solutions.
(a) 0.300 M propanoic acid (HC3H5O2, Ka = 1.3 10-5)
(b) 0.300 M sodium propanoate (NaC3H5O2)
(c) pure H2O
(d) 0.300 M HC3H5O2 and 0.300 M NaC3H5O2

  • chemistry -

    a)Let HA = the weak acid, proanoic acid.
    ..............HA ==> H^+ + A^-
    initial......0.300....0......0
    change.........-x......x......x
    equil........0.300-x...x.......x

    Ka = (H^+)(A^-)/(HA)
    Substitute from the ICE chart and solve for x, then convert to pH.

    b)
    This is a salt and the pH will be determined by the hydrolysis of the salt.
    A^- + HOH ==> HA + OH^-
    Set up an ICE chart similar to the one in (a) and solve for OH^-, then convert to pH.

    (c) Pure water ionizes as
    HOH => H^+ + OH^-
    Set up an ICE chart and use
    Kw = (H^+)(OH^-), solve for H^+ and convert to pH.

    (d)
    This is a buffered solution. Use the Henderson-Hasselbalch equation.
    Post your work if you get stuck.

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