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16 October, 08:07

Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5 * 10-4.

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  1. 16 October, 08:11
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    The pH of the solution after adding 0.150 moles of solid LiF is 3.84

    Explanation:

    We have the chemical equation,

    HF (aq) + NaOH (aq) - >NaF (aq) + H2O

    To find how many moles have been used in this

    c = n/V=> n = c. V

    nHF=0.250 M⋅1.5 L=0.375 moles HF

    Simillarly

    nF=0.250 M⋅1.5 L=0.375 moles F

    nHF=0.375 moles - 0.250 moles=0.125 moles

    nF=0.375 moles+0.250 moles=0.625 moles

    [HF]=0.125 moles/1.5 L=0.0834 M

    [F-]=0.625 moles/1.5 L=0.4167 M

    To determine the problem using the Henderson - Hasselbalch equation

    pH=pKa+log ([conjugate base/[weak acid])

    Find the value of Ka

    pKa=-log (Ka)

    pH=-log (Ka) + log ([F-]/[HF]

    pH = - log (3.5 x 10 ^4) + log (0.4167 M/0.0834 M)

    pH=-log (3.5 x 10 ^4) + log (4.996)

    pH = - 4.54+0.698

    pH = - (-3.84)

    pH=3.84

    The pH of the solution after adding 0.150 moles of solid LiF is 3.84
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