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Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 2.15 M...

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Chemistry

Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mLof 2.15 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of the bufferafter 0.750 mol of H⁺ have been added?

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4.2 Ratings (882 Votes)

no of moles of NaC7H5O2   = W/G.M.Wt

                                       = 132.8/144   = 0.923moles

no of moles of HC7H5O2   = molarity * volume in L

                                      = 2.15*0.3   = 0.645moles

PKa = -logKa

         = -log(6.3*10^-5)

         = 4.2

no of moles of NaC7H5O2 after adding 0.750 mol of H^+ = 0.923-0.75   = 0.173 moles

no of moles of HC7H5O2 after adding 0.750mol of H^+ = 0.645+0.75    = 1.395moles

PH   = Pka + log[NaC7H5O2]/[HC7H5O2]

      = 4.2 + log0.173/1.395

      = 4.2 -0.9065

       = 3.2935>>>>answer
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