Question

Determine the pH of each of the following solutions.

(a) 0.113 M ascorbic acid (weak acid with K_a = 8e-05).





(b) 0.261 M phenol (weak acid with K_a = 1.3e-10).





(c) 0.150 M pyridine (weak base with K_b = 1.7e-09).

Answer 1

(a)

Ka = 8.0 × 10^-5 and c = 0.113M

For weak acids the relation for pH can be given as:

pH = 1/2[pKa - logc] ------------------(1)

Where, in this case pKa of ascorbic acid = -logKa = -log8.0 x 10^-5

pKa = - (-4.09)

pKa = 4.09

put all these values in above equation (1):

pH = 1/2[4.09 - log0.113]

pH = 1/2[4.09 –(-0.95)]

pH = 1/2[4.09 + 0.95)]

pH = ½[5.04]

pH = 2.52


(b)

Ka = 1.3 × 10^-10 and c = 0.261M

From similar relation:

pH = 1/2[pKa - logc] ------------------(1)

Where, in this case pKa of phenol = -logKa = -log1.3 x 10^-10

pKa = - (-9.88)

pKa = 9.88

put all these values in above equation (1):

pH = 1/2[9.88 - log0.261]

pH = 1/2[9.88 –(-0.58)]

pH = 1/2[9.88 + 0.58)]

pH = ½[10.46]

pH = 5.23

(c)

Kb = 1.7 x 10^-9 and concentration c = 0.15M

pH for weak base can be determined by relation:

pH = 14 - 1/2[pKb - logc] ------------------(1)

Where, in this case pKb of pyridine can be obtained as:

pKb = -(log1.7 x 10^-9)

pKb = -(-8.77)

pKb = 8.77

put all these values in above equation (1):

pH = 14 - 1/2[8.77 - log0.15]

pH = 14 - 1/2[8.77 –(-0.82)]

pH = 14 - 1/2[8.77 + 0.82)]

pH = 14 - ½[9.59]               

pH = 14 – 4.765

pH = 9.20