Use the systematic treatment of equilibrium to determine the pH and concentratio

Question

Use the systematic treatment of equilibrium to determine the pH and concentrations of species in 1.00 L of solution containing 0.070 mol lysine (HL), 0.055 mol aspartic acid (H2A), and 0.0080 mol NaOH. Consider just acid-base chemistry. Ignore ion pairing and activity coefficients. Lysine (HL) is derived from the triprotic acid H3L2 with pKH3L2 = 1.77, pKH2L = 9.07, and pKHL = 10.82. Aspartic acid (H2A) is derived from the triprotic acid H3A with pKH3A = 1.990, pKH2A = 3.900, and pKHA– = 10.002.

Explanation / Answer

Formulae used :

pKa = - log Ka    so if p Ka = 5 then   Ka = 10^-5

ph = - Log(H+ ion conc )

For LYSINE

H+ ion conc = ( Ka1*C ) ^1/2 + ( Ka2*C ) ^1/2 + ( Ka3*C ) ^1/2

                  =( C ) ^1/2(( Ka1 ) ^1/2+( Ka2 ) ^1/2 + (Ka3 ) ^1/2)

                  = ( 0.070 ) ^1/2((10^-10.82 ) ^1/2 +( 10^-9.07 ) ^1/2 +( 10^-1.77 ) ^1/2)

                  = o.2646 (3.89*10^-6 + 2.917*10^-5 + 0.13032)

                   = 0.03449M      -------------------------(1)

For Aspartic acid

H+ ion conc = ( Ka1*C ) ^1/2 + ( Ka2*C ) ^1/2 + ( Ka3*C ) ^1/2

                  =( C ) ^1/2(( Ka1 ) ^1/2+( Ka2 ) ^1/2 + (Ka3 ) ^1/2)

                =   ( 0.055 ) ^1/2(( 10^-10.002 ) ^1/2 + ( 10^-3.900 ) ^1/2 + ( 10^-1.990 ) ^1/2)

               = 0.2345(9.977*10^-6 + 0.01122 + 0.98829)

                = 0.2344

For mixture of LYSINE and Aspartic acid :

    H+ ion conc - OH- ion conc = (0.03449+0.2344) - (0.0080) = 0.26089 (because OH - conc = conc of NaOH)

pH = - log(H+ ion conc) = - log(0.26089) = 0.58354

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