Question

a monoprotic weak acid HB has an acid ionization constant Ka of 5.920x10^-10. Using only .300 M HB and 0.300 M NaB how would you prepare 150ml of a buffer having a pH of 9.65?

Answer 1

Suppose we add V mL of HB.

Total Volume of buffer = 150 mL

=> Volume of NaB added = 150 - V

Moles = Molarity x Volume (L)

=> Moles of HB added = 0.3 x V = 0.3 V mmol

Moles of NaB added = 0.3 x (150 - V) = (45 - 0.3 V) mmol

pH = pKa + log (NaB / HB)

Ka = 5.92 x 10^-10

=> pKa = 9.228

=> 9.65 = 9.228 + log [ (45 - 0.3 V) / (0.3 V) ]

=> log [ (45 - 0.3 V) / (0.3 V) ] = 0.422

=> [ (45 - 0.3 V) / (0.3 V) ] = 2.64

=> V = 41.21 mL

Theerfore,

Volume of 0.3 M HB added = 41.21 mL

and Volume of 0.3 M NaB added = 108.79 mL