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

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

Expert Solution

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

queen_honey_blossom answered 2 years ago

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