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When 22.50 mL of aqueous NaOH were added to 1.311 g of cyclohexylaminoethanesulfonic acid (FM 207.29,...

When 22.50 mL of aqueous NaOH were added to 1.311 g of cyclohexylaminoethanesulfonic acid (FM 207.29, structure in the pKa values of common buffers table) dissolved in 41.18 mL of water, the pH was 9.25. Calculate the molarity of the NaOH.

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Answer #1

pH of acidic buffer = pka + log(NaOH/cyclohexylaminoethanesulfonic acid)

no of mole of NaOH = x

no of mole of cyclohexylaminoethanesulfonic acid = 1.311/207.29 = 0.00632 mol

pka of cyclohexylaminoethanesulfonic acid = 9.3


pH = 9.25


   9.25 = 9.3 +log(x/(0.00632-x))

x = 0.00298


no of mole of NaOH = x = 0.00298 mol

concentration of NaOH added = n/v*1000

                           = 0.00298/22.5*1000

                           = 0.1324 M

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