Question

Determine the pH of each of the following solutions.

(a) 0.637 M hydrosulfuric acid (weak acid with K_a = 9.5e-08).





(b) 0.764 M hypoiodous acid (weak acid with K_a = 2.3e-11).





(c) 0.458 M pyridine (weak base with K_b = 1.7e-09).

Answer 1

(a) Given concentration "c" = 0.637 M hydrosulfuric acid and weak acid with Ka = 9.5 x 10-08.

For weak acids (no 100% ionization) the relation for pH can be given as:

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

where in this case pKa of hydrosulfuric acid = -logKa = -log9.5 x 10^-8

pKa = - (-7.02)

pKa = 7.02

put all these values in above equation (1):

pH = 1/2[7.02 - log0.637]

pH = 1/2[7.02 –(-0.196)]

pH = 1/2[7.02 + 0.196)]

pH = ½[7.216]

pH = 3.61

(b) given "c" = 0.764 M hypoiodous acid and weak acid with Ka = 2.3 x10^-11

pKa of hypoiodous acid = -logKa = -log2.3 x 10^-11

pKa = - (-10.64)

pKa = 10.64

Now, put all these values in above equation (1):

pH = 1/2[10.64 - log0.764]

pH = 1/2[10.64 –(-0.117)]

pH = 1/2[10.64 + 0.117)]

pH = ½[10.76]

pH = 5.38

(c) Given "c" = 0.458 M pyridine and weak base with Kb = 1.7 x 10^-09.

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 - log0.458]

pH = 14 - 1/2[8.77 –(-0.339)]

pH = 14 - 1/2[8.77 + 0.339)]

pH = 14 - ½[9.109]               

pH = 14 – 4.55

pH = 9.44