Answer:
5.42
Explanation:
Step 1: Consider the dissociation of NH₄Br
NH₄Br(aq) ⇒ NH₄⁺(aq) + Br⁻(aq)
Br⁻ is the conjugate base of HBr, a strong acid, so it doesn´t react with water. NH₄⁺ is the conjugate acid of NH₃, so it does react with water.
Step 2: Consider the acid reaction of NH₄⁺
NH₄⁺(aq) + H₂O(l) ⇄ NH₃(aq) + H₃O⁺(aq)
Step 3: calculate the acid dissociation constant for NH₄⁺
We will use the following expression.
[tex]K_a \times K_b = K_w\\K_a = \frac{K_w}{K_b} = \frac{1.00 \times 10^{-14} }{1.76 \times 10^{-5}} = 5.68 \times 10^{-10}[/tex]
Step 4: Calculate the concentration of H₃O⁺
We will use the following expression.
[tex][H_3O^{+} ]= \sqrt{K_a \times C_a } = \sqrt{5.68 \times 10^{-10} \times 0.0255 } = 3.81 \times 10^{-6}M[/tex]
Step 5: Calculate the pH
We will use the following expression.
[tex]pH = -log [H_3O^{+} ] = -log (3.81 \times 10^{-6}) = 5.42[/tex]
The pH of 0.0255 M solution should be 5.42.
Since we know that
ka * kb = kw
So,
ka = kw/kb
= 1.00*10^-14 / 1.76*10^-5
= 5.68*10^-10
Now the concentration of H3O should be
= √ka * Ca
= √5.68*10^-10 * 0.0255
= 3.81*10^-6M
Now the pH value should be
= -log(H3O+)
= -log(3.81*10^-6)
= 5.42
hence, The pH of 0.0255 M solution should be 5.42.
Learn more about pH here: https://brainly.com/question/23506014
