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Chemistry

A 100 mL sample of 0.40 M hydrofluoric acid is mixed with 100.0 mL of 0.2M lithium hydroxide. Find the pH of the final solution. pKa for HF is 3.2.

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  1. millimolwa HF = mL x M = 100 mL x 0.40 M = 40
    millimoles LiOH = 100 x 0.2 M = 20
    ..................HF + LiOH ==> LiF + H2O
    I................40..........0.............0.........0
    add........................20............................
    C...............-20.......-20............+20..................
    E...............20..........0...............20.......................
    So you have formed a buffer solution of HF and F (a weak acid and its salt in solution is a buffere). Use the Henderson-Hasselbalch equation to solve for pH. That's pH = pKa + log [(base)/(acid)]
    You will need pKa. pKa = -log Ka for HF. You can find Ka in your text or on the webl.
    You will need concn base (F^-) = 20 mmols/200 mL = ?
    You will need concn acid (HF) = 20 mols/200 mL = ?
    Post your work if you get stuck.

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    DrBob222
  2. Would the equation for pH be pH=3.2+log[0.01/10] ?
    (using the given pKa and molarity of the base and acid)

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  3. yes and no.
    pH = pKa + log (F^-)/(HF)
    pKa is 3.2---yes
    But your concentrations are not right.
    (F^-) = 20 millimols/200 mL = 0.1
    (HF) = 20 millimoles/200 mL = 0.1
    pH = 3.2 + log (0.1/0.1) = 3.2 + log 1 = 3.2 + 0 = 3.2

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    DrBob222

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