Question

A buffer solution contains 0.56 mol of ascorbic acid (HC6H7O6) and 0.54 mol of sodium ascorbate (NaC6H7O6) in 5.70 L. The Ka of ascorbic acid (HC6H7O6) is Ka = 8e-05.

(a) What is the pH of this buffer?

(b) What is the pH of the buffer after the addition of 0.10 mol of NaOH? (assume no volume change)

(c) What is the pH of the original buffer after the addition of 0.30 mol of HI? (assume no volume change)

Answer 1

we can use mol instead of concentration since volume will be canceled out in numerator and denominator

a)

pka = -log ka

     = -log (8*10^-5)

     = 4.097

use:

pH = pKa + log {[ NaC6H7O6]/[ HC6H7O6]}

    = 4.097 + log (0.54/0.56)

    =4.081

b)

adding 0.10 mol NaOH will increase number of moles of NaC6H7O6 by 0.10 mol and decrease mol of HC6H7O6 by0.10 mol

pH = pKa + log {[ NaC6H7O6]/[ HC6H7O6]}

    = 4.097 + log ((0.54+0.10)/(0.56-0.10))

    =4.240

c)

adding 0.30 mol HI will increase number of moles of HC6H7O6 by 0.30 mol and decrease mol of NaC6H7O6 by0.30 mol

pH = pKa + log {[ NaC6H7O6]/[ HC6H7O6]}

    = 4.097 + log ((0.54- 0.30)/(0.56 + 0.30))

    = 3.543