Question

Calculate the pH of each of the following solutions. a. 0.100 M propanoic acid (HC3H5O2, Ka 1.3 105 ) b. 0.100 M sodium propanoate (NaC3H5O2) c. pure H2O d. a mixture containing 0.100 M HC3H5O2 and 0.100 M NaC3H5O2

Answer 1

Answer:

a) pH = 2.95

b) pH = 8.94

c) pH = 7.00

d) pH = 4.89

Explanation:

a) The reaction is:

               CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)(1)

At equlibrium:              0.1 - x                                     x                       x

The dissociation constant of equation (1), Ka, is:

$K_{a} = \frac{[C_{3}H_{5}O_{2}^{-}][H_{3}O^{+}]}{[C_{3}H_{6}O_{2}]}$

$1.3\cdot 10^{-5} = \frac{x^{2}}{(0.1 - x)}$   (2)

Solving equation (2) for x, we have:

x = 0.00113 = [H₃O⁺] = [CH₃CH₂COO⁻]      

Hence, the pH is:

$pH = -log [H_{3}O^{+}] = -log (0.00113) = 2.95$

b) The reaction is:  

            CH₃CH₂COO⁻(aq) + H₂O(l)  ⇄ CH₃CH₂COOH(aq) + OH⁻(aq)

At equlibrium:    0.1 - x                                        x                         x  

The dissociation constant of the above equation, Kb, is:  

$K_{b} = \frac{[C_{3}H_{6}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}$

Also, we have that:

$K_{w} = K_{a}*K_{b} \rightarrow K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \cdot 10^{-14}}{1.3 \cdot 10^{-5}} = 7.69 \cdot 10^{-10}$

$7.69\cdot 10^{-10} = \frac{x^{2}}{0.1 - x}$    (3)

Solving equation (3) for x, we have:

x = 8.77x10⁻⁶ = [OH⁻] = [CH₃CH₂COOH]

Hence, the pH is:

$pOH = -log[OH^{-}] = -log(8.77\cdot 10^{-6}) = 5.06 \rightarrow pH = 14 - pOH = 8.94$    

c) Pure water:

H₂O ⇄ H⁺ + OH⁻

$K_{w} = [H^{+}][OH^{-}] \rightarrow 1\cdot 10^{-14} = [H^{+}][OH^{-}]$

Since is pure water: [H⁺] = [OH⁻]  

$1\cdot 10^{-14} = [H^{+}]^{2} \rightarrow [H^{+}] = \sqrt{1\cdot 10^{-14}} = 1 \cdot 10^{-7}$

Therefore, the pH is:

$pH = -log [H^{+}] = -log (1 \cdot 10^{-7}) = 7.00$

d) The reaction is:

          CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)  

At equlibrium:           0.1                                     0.1                        x

The pH is:

$pH = pKa + log(\frac{[C_{3}H_{5}O_{2}^{-}]}{[C_{3}H_{6}O_{2}]}) = -log(1.3 \cdot 10^{-5}) + log(\frac{0.1}{0.1}) = 4.89$

I hope it helps you!