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Analytical chemistry help!

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Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN (Ka of HCN = 6.2 x 10-10) with a) 4.20 mL of 0.438 M HClO4 and b) 11.82 mL of 0.438 M HClO4. What is the pH at the equivalence point with 0.438 M HClO4?

  • Analytical chemistry help! - ,

    The data:
    millimoles NaCN = 50.00*0.100 = 5.00
    mmoles HClO4 = 4.20*0.438 = 1.8396 (I know that's too many significant figures but you can always drop the extra ones at the end.)
    mmoles HClO4 = 11.82*0.438 = 5.177
    ----------------------------------------
    I'm not copying all of the zeros; they sometimes mess up the spacing on this ICE chart.
    .........NaCN + HClO4 ==> HCN + NaClO4
    initial.. 5......0.........0.......0
    add...........1.8396................
    change..-1.8396..-1.8396...1.8396..etc
    equil....3.16...0..........1.8396...etc

    You can see for part a that you have a weak acid (HCN) and a salt of the weak acid in solution (NaCN) so this is a buffered solution. Use the Henderson-Hasselbalch equation and solve for pH.

    part b. Set up an ICE chart as in part a to see what you have. I think you will have 0.177 mmoles HClO4 remaining in soln along with 5 mmoles HCN. So that is a weak acid plus a common ion (H^+) from the HClO4. You make another ICE chart with HCN==> H^+ + CN^-
    substitute into Ka (remember to add the H^+ common ion), and solve for pH.

    The last part for pH at equivalence point, that is determined by the pH of the NaCN salt. Set up the hydrolysis and use an ICE chart for that. You will need to determine the (CN^-) at the equivalence point which I will call c.
    ........CN^- + HOH ==> HCN = OH^-
    .........c...............0.....0
    change...-x.............x.......x
    equil....c-x............x.......x

    Kb = (Kw/Ka) = (HCN)(OH^-)/(CN^-)
    Kw = 1E-14
    Ka = Ka for HCN
    (HCN) + (OH^-) = x
    (CN^-) = c-x
    Solve for x = OH^- and convert to pH.
    Post your work if you get stuck.

  • Analytical chemistry help! - ,

    I converted every thing to mols instead of milli mols so for the first part got

    pH=9.208 + log(1-(4.20/11.42)/(4.20/11.42)

    so my pH= 9.44 when 4.20 mL of 0.438 M HClO4 is added?

  • Analytical chemistry help! - ,

    That looks good to me.

  • Analytical chemistry help! - ,

    Great I think I can figure the rest out! Thank you for your help!

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