Question

Determine the pH of each of the following solutions. (a) 0.571 M carbonic acid (weak acid with Ka = 4.3e-07). (b) 0.332 M propionic acid (weak acid with Ka = 1.3e-05). (c) 0.308 M pyridine (weak base with Kb = 1.72-09).

Answer 1

of the acid Determine the pH of each following solutions a) 0.571M carbonic acid (weak Ka= 4.30-07) Answer = (pH = 3.30 solution For calculating pH of weak acid, first we calculate degree of dissociation (@) by using following formula ka = ca ² |-(i) where Ka dissociation constant c = concentration a = degree of dissociation Ka for carbonic acid = 4.30-07= 4.3x10 Concentration of acid = C = 0.571 M i we calculate value of a using equation (1)

4.38 10 7 = 0.571 XQ² x = 4.3 x 10+ 0.571 q? = 7:53 x 10 x ² = 75.3 x 10 Q = 575.3 x 10 8 Q = 8.67X 104 After calculating a, we can calculate [ht] ion concentration using following formula [[ht] =ca] (ii) but the value of C && then calculate value of [H+] [H+] = 0.571 X8.67 10 4 [H+] = 4.95% 10 ) After calculating [ht], we can calculate pH by the following formula - can

PH = -log[H+] - (iii) but the value of [H + ] in equation in). pH = -log(4.95 x 10-4) PH= 3.30 a for Carbonic acid Answer=[pH = 3.30 (weak acid 0.332 M propanoic acid Ka = 1:30-05) Answer = PH = 2.687 solution. This problem is similar to brevious problem. So we follow same procedure for calculating pH. first we calculate value of a using following equation, [ka = (42) - (1) value of Ka= 1.3e-05 = 1.38 10 -5

Concentration of acid = 0.332 M so but these value in equation (i). 1.38 10 5 = 0.332 & 2 x2 = 1.38 10 3 0.332 q? = 3.915 x 10-5 22- 39.15 x 10 Q = 539:15 X 20-6 x = 6.25 x 103 | After calculating & , we can calculate [Ht ] using following formula. [H+] = Ca] - Lil but the value of C && in equation (ii) [H+] = 0.332 x 6.25 X 10 - 3 [H+] = 2.075 x 10 3 After calculating [H + ], we can calculate pH of weak acid by following formula.

|PH = -log [H+] - (ii) but the value of [H+j in equation (ii) pH = -log C2.075 x 10-3) [pH = 2.68] Answer=[pH of bropanic acid=2.68 base c) 0.308 M pyridine (weak eta kb= 1.72-09) Answert solution. For calculating pH of weak base, we first calculate value of a by following relation, [kg = CQ² - (i) where Kb = dissociation constant of weak bese = concentration of weak base a = degree of dissociation

Kb value of weak base = 1.7e-og=1.78 10 3 concentration of base = 0.300M on but these value in equation (is. 1.7X10-9 - 0.308 xq? -9 L = 1.7 X 10 0.300 22= 5:519 x 10-9 d = 55.19 x 10-10 = 155-19 x 10-10 Q = 7.42 x 10-5 After calculating &, we can calculate [OH- ] for weak base using This formula . [OH-] = ca - (11) but the value in equation (ii) TOH ] = 0.308 X 7.42 x 105 [OH-] = 2.28 x 10-51

After calculating [OH-], we can calculate poH by following formula TPOH = -log [OH-] - (iii) but the value of [OH ] in equation ilin pOH = -log (2.20x10-5) IPOH = 4.647 After calculating pOH , we can calculate ph by following formula. [PH + pOH = 14.0) – civy but the value of poh and calculate value of pH .. pH + 4.64 = 14.00 pH = 14.00 -4.64 pH=9.36 Answer = pH of pyridine = 9.36
summary:

For calculating pH of weak acid , we simply used following formula

Ka = c × (degree of dissociation)^2

[H+] = c × degree of dissociation

pH = -log [H+]

Where

Ka = dissociation constant of weak acid

For calculating pH of weak base , we used following formula

Kb= c × (degree of dissociation)^2

[OH-] = c × degree of dissociation

pOH = -log [OH-]

pH + pOH= 14

Where

Kb = dissociation constant of weak base

c = concentration of base