What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid...

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What is the pH of 1.000 L of a solution of 100.0 g of glutamicacid (C5H9NO4, a diprotic acid; K1 = 8.5 ×
10−5, K2 = 3.39 × 10−10) to which has been added 20.0 g of NaOHduring the preparation of monosodium
glutamate, the flavoring agent? What is the pH when exactly 1 molof NaOH per mole of acid has been added?

Answer & Explanation Solved by verified expert
4.1 Ratings (552 Votes)

no of moles    = W/G.M.Wt

              = 100/147*1    = 0.68 M

    no of moles of NaOH = W/G.M.Wt   = 20/40 = 0.5 moles

         C5H9NO4 + NaOH --------> NaC5H8NO4 + H2O

I      0.68                0.5                     0   

C   -0.5               -0.5                      0.5

E   0.18              0                          0.5

    Pka = -logKa

           = -log8.5*10-5

          = 4.0705

PH    = Pka + log[NaC5H8NO4]/[C5H9NO4]

         = 4.0705 + log0.5/0.18

         = 4.0705 + 0.4436

         = 4.5141

     


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