Question

Five milliliters (5.0 mL) of 0.100 M NaOH is added to 145 mL of a 0.20 M Tris/TrisH+ buffer at pH 8.30. Is the resulting solution still a good buffer, i.e., what is the resulting pH? The pKa of TrisH+ is 8.30.

Answer 1

Given data,

pH = 8.30

pKa of TrisH+ is 8.30

total buffer = V * M

= 0.145 * 0.2

= 0.029 mol

Let us consider,

pH = 14 - ( pkb + log(TrisH+/Tris base))

Since, pka = 14 - pkb

pH = pka - log(TrisH+/Tris base))

8.3 = 8.3 - log (x / (0.029-x))

0 = -  log (x / (0.029-x))

x / (0.029-x) = 10^-0

x = 0.0145

No of mol of TrisH+ = x = 0.0145 mol

No of mol of Tris = 0.029-0.0145

= 0.0145 mol

after addition of NaOH

pH = pka - log(TrisH+ - NaOH/Tris+NaOH))

concentration of NaOH added = 5 * 0.1/ 150

= 0.0033 M

pH = 8.3 - log((0.0145-0.0033) / (0.0145 + 0.0033))

= 8.3 - log( 0.0112 / 0.0178)

= 8.3 + 0.20

pH = 8.5

buffer capacity, from (pka + 1) to (pka - 1).

so that, it is still good buffer.