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  3. a 1.50 l buffer solution is 0.250 m in hf and 0.25

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate...

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Chemistry

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF.Calculate the pH of the solution after the addition of 0.0500 molesof solid NaOH. Assume no volume change upon the addition of base.The Ka for HF is 3.5 × 10-4.​

Answer & Explanation Solved by verified expert
3.8 Ratings (526 Votes)

no of moles of HF = molarity * volume in L

                            = 0.25*1.5   = 0.375 moles

no of moles of NaF = molarity * volume in L

                               = 0.25*1.5   = 0.375 moles

By the addition of NaOH

no of moles of HF = 0.375-0.05   = 0.325 moles

no of moles of NaF = 0.375 + 0.05 = 0.425 moles

    Pka = -logKa

             = -log3.5*10-4   = 3.4559

PH   = PKa + log[NaF]/[HF]

         = 3.4559 + log0.425/0.325

         = 3.4559 + 0.1165   = 3.5724
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