Question

A buffer solution contains 0.29 mol of phenol (HC₆H₅O) and 0.75 mol of sodium phenoxide (NaC₆H₅O) in 5.20 L.
The K_a of phenol (HC₆H₅O) is K_a = 1.3e-10.



(a) What is the pH of this buffer?

pH = ?????????????


(b) What is the pH of the buffer after the addition of 0.06 mol of NaOH? (assume no volume change)

pH = ????????????


(c) What is the pH of the original buffer after the addition of 0.59 mol of HI? (assume no volume change)

pH = ????????????

Answer 1

'Henderson-Hasselbalch' equation is

pH = pKa + log[conj.base/conj.acid]

(a) pKa = -log[Ka] = -log (1.3x 10^-10) = 9.88

Here Phenol is the acid and sodium phenolxide is the base

0.29 moles of phenol in 5.2 L will be 0.056 M

0.75 moles of sodium phenoxide in 5.2 L will be 0.144 M

pH = pKa + log[conj.base/conj.acid]

pH = 9.88 + log [0.144/0.056]

pH = 9.88 + log (2.57)

pH = 9.88 +0.4099

pH = 10.29

b) What is the pH of the buffer after the addition of 0.06 mol of NaOH? (assume no volume change)

NaOH + C₆H₅OH (phenol) ----> NaC₆H₅O (sodium phenoxide) + H₂O

on adding 0.06 moles of NaOH, phenol is reduced to 0.23 moles = 0.044 M

on adding 0.06 moles of NaOH, sodium phenoxide is increased to 0.81 moles = 0.156 M

pH = pKa + log[conj.base/conj.acid]

pH = 9.88 + log [0.156/0.044]

pH = 9.88 + log (3.54)

pH = 9.88 +0.549

pH = 10.43

(c) What is the pH of the original buffer after the addition of 0.59 mol of HI? (assume no volume change)

Na(C₆H₅O) (sodium phenoxide) + HI -------> C₆H₅OH (phenol) + NaI

on adding 0.59 moles of HI sodium phenoxide is consumed to and 0.16 moles of sodium phenoxide remains = 0.0307 M

on adding 0.59 moles of HI, phenol increses to give 0.88 moles = 0.169 M

pH = pKa + log[conj.base/conj.acid]

pH = 9.88 + log [0.0307/0.169]

pH = 9.88 + log (0.181)

pH = 9.88 - 0.74

pH = 9.139