Question

9-19. Calculate the pH of water at 25°C and 75°C. The values for pKw at these temperatures are 13.99 and 12.70, respectively. ​

Answer 1

Answer:

7.00; 6.35

Explanation:

1. At 25 °C

2H₂O ⇌ H₃O⁺ + OH⁻

pH = ½pKw = ½ × 13.99 = 7.00

2. At 75° C

pH = ½pKw = ½ × 12.70 = 6.35

Answer 2

The values of the pH of water at 25 °C (pKw = 13.99) and 75 °C (pKw = 12.70) are 6.99 and 6.35, respectively.

The equation for water dissociation is the following:

2H₂O ⇄ H₃O⁺ + OH⁻

The dissociation constant of the above reaction is:

$K_{w} = [H_{3}O^{+}]*[OH^{-}]$

Since the solution has only water, the concentrations of H₃O⁺ and OH⁻ are equal, so:

$K_{w} = [H_{3}O^{+}]^{2}$      

$[H_{3}O^{+}] = \sqrt{K_{w}}$   (1)

Now, the pH given by:

$pH = -log([H_{3}O^{+}])$   (2)

So, by entering equation (1) into (2):

$pH = -log([H_{3}O^{+}]) = -log(\sqrt{K_{w}})$   (3)

The pKw is equal to:

$pK_{w} = -log(K_{w})$

$K_{w} = 10^{-pK_{w}}$   (4)

Now, entering eq (4) into (3):

$pH = -\frac{1}{2}log(K_{w}) = \frac{1}{2}pK_{w}$   (5)  

1. For the temperature of 25°C we have:

$pH = \frac{1}{2}pK_{w} = \frac{1}{2}13.99 = 6.99$      

Hence, the pH of water at 25 °C is 6.99.

2. For the temperature of 75°C we have:

$pH = \frac{1}{2}12.70 = 6.35$      

Then, the pH of water at 75 °C is 6.35.

Therefore, the pH of water at 25 °C and 75 °C are 6.99 and 6.35, respectively.

You can find more about pH here: https://brainly.com/question/491373?referrer=searchResults

I hope it helps you!