A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 3.00 mmol of the solid acid in 100.0 mL water and titrates the resulting solution with 0.00500 M NaOH. After 10.0 mL NaOH has been added, the pH is 4.88. What is the Ka value for the acid?

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answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole