Question

A buffer solution contains 0.54 mol of boric acid (H₃BO₃) and 0.89 mol of sodium dihydrogen borate (NaH₂BO₃) in 4.80 L.
The K_a of boric acid (H₃BO₃) is K_a = 5.8e-10.



(a) What is the pH of this buffer?

pH =  


(b) What is the pH of the buffer after the addition of 0.18 mol of NaOH? (assume no volume change)

pH =  


(c) What is the pH of the original buffer after the addition of 0.12 mol of HI? (assume no volume change)

pH =  

Answer 1

a) for calculating pH of acidic buffer we use following equation.

pH = pKa + log [sodium borate] / [boric acid]

pKa = - log Ka = - log [5.8 x 10^-10]

pKa = 9.24

[boric acid] = 0.54 / 4.80 = 0.1125 M

[sodium borate] = 0.89 / 4.80 = 0.1854 M

pH = 9.24 + log [0.1854] / [0.1125]

pH = 9.42

b) after addition of 0.18 / 4.8 = 0.0375 M NaOH added

[boric acid] = 0.1125 - 0.0375 = 0.075 M

[sodium borate] = 0.1854 + 0.0375 = 0.2229 M

pH = 9.24 + log [0.229] / [0.075]

pH = 9.72

c) after addition of 0.12 / 4.80 = 0.025 M HI

[boric acid] = 0.1125 + 0.025 = 0.1375 M

[sodium borate] = 0.1854 - 0.025 = 0.1604 M

pH = 9.24 + log [0.1604] / [0.1375]

pH = 9.099.30