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1 June, 03:30

A 1.50L buffer solution is 0.250 M in HF and 0.250 in NaF. Calculate the ph of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of the base. The Ka for HF is 3.5E^-4

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  1. 1 June, 03:59
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    The pH is 3.57

    Explanation:

    Step 1: Data given

    Volume = 1.50 L

    Molarity of HF = 0.250 M

    Molarity of NaF = 0.250 M

    addition of 0.05 moles NaOH

    Step 2:

    The buffer contains HF, which is a weak acid; and NaF, which is the salt of its conjugate base, the fluoride anion, F.

    -

    When adding NaOH (and colume will not change), we expect the weak acid to neutralize the strong base (and vice versa).

    HF (aq) + NaOH (aq) → NaF (aq) + H2O (l)

    For 1 mole HF consumed, we need 1 mole NaOH, to produce 1 mole NaF and 1 mole H2O

    Step 3: Calculate moles of HF

    Moles HF = Molarity HF * Volume HF

    Moles HF = 0.250 M * 1.5 L

    Moles HF = 0.375 moles HF

    Step 4: Calculate moles of F - (conjugate base)

    Moles F - = 0.250 M * 1.5 L

    Moles F - = 0.375 moles F-

    Step 5: Addition of 0.05 moles NaOH

    HF + OH - → F - + H2O

    Since we have fewer moles of strong base than on weak acid, the sodium hydroxide will be completely consumed.

    The number of moles HF will change to:

    Moles HF = 0.375 - 0.05 = 0.325 moles HF

    The number of moles F - will change to:

    Moles F - = 0.375 + 0.05 = 0.425 moles F-

    Step 6: Calculate molarity

    Molarity HF = Moles HF / Volume

    Molarity HF = 0.325/1.5 = 0.21667 M

    Molarity F - = 0.425/1.5 = 0.28333 M

    Step 7: Calculate pKa

    3.5 * 10^-4 = Ka pKa = - log (3.5 * 10^-4) = 3.456

    Step 8: Calculate pH

    pH = - log (3.5 * 10^-4) + log[F-]/[HF]

    pH = 3.456 + log (0.28333/0.21667)

    pH = 3.57

    The pH is 3.57
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