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Posted by on Saturday, January 21, 2012 at 11:17pm.

A solution that contains 2.047 x 10^-1 M of acid, HA, and 1.808 x 10^-1 M of its conjugate base, A-, has a pH of 4.196. What is the pH after 1.629 x 10^-3 mol NaOH is added to 5.961 x 10^-1 L of this solution?

  • Chemistry - , Saturday, January 21, 2012 at 11:41pm

    (HA) = 0.2047
    (A^-) = 0.1808
    moles HA = 0.2047M*0.5961L = 0.12202
    moles A^- = 0.1808 x 0.5961 = 0.10777

    pH = pKa + log (base)/(acid)
    The problem states the pH as well as the beginning HA and A. Substitute those and solve for pKa. Then look at the ICE chart below.
    ............HA + OH^- ==> A^- + H2O
    initial.0.12202...0...0.10777....
    add............0.001629............
    change..-.001629.-.001629..+.001629.....
    equil......?......0.......?
    Fill in the ? spots and substitute into the Henderson-Hasselbalch equation to solve for new pH. Use pKa from the first calculation above. Post your work if you need additional help. I have carried more digits than allowed; you should round as needed.

  • Chemistry - , Saturday, January 21, 2012 at 11:54pm

    What would the Ka for this be?

  • Chemistry - , Saturday, January 21, 2012 at 11:58pm

    Oops, sorry. I understand now, thank you.

  • Chemistry - , Saturday, January 21, 2012 at 11:59pm

    I calculated pKa = 4.25 so Ka = 5.624E-5

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