Respuesta :
Explanation:
The given data is as follows.
[tex][P^{2-}][/tex] = 0.042 M, [tex]K_{a}[/tex] for [tex]HP^{-} = 3.9 \times 10^{-6}[/tex]
According to the given situation [tex]P^{2-}[/tex] acts as a base.The reaction equation will be as follows.
[tex]P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}[/tex]
Relation between [tex]K_{b}[/tex] and [tex]K_{a}[/tex] are as follows.
[tex]K_{a} \times K_{b} = K_{w}[/tex]
[tex]K_{b} = \frac{1 \times 10^{-14}}{K_{a}}[/tex]
= [tex]\frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}[/tex]
= [tex]2.6 \times 10^{-9}[/tex]
Also, [tex]K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}[/tex]
Let us take [tex][OH^{-}] = [HP^{-}][/tex] = x
So, [tex]K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}[/tex]
[tex]2.6 \times 10^{-9} = \frac{x \times x}{0.042}[/tex]
x = [tex]1.04 \times 10^{-5}[/tex]
[tex][OH^{-}] = [HP^{-}][/tex] = [tex]1.04 \times 10^{-5}[/tex]
pOH = - log[tex][OH^{-}][/tex]
= - log ([tex]1.04 \times 10^{-5}[/tex])
= 4.99
As it is known that pH + pOH = 14
so, pH + 4.99 = 14
pH = 9.01
Thus, we can conclude that pH of the solution is 9.01.
The pH of the solution will equal 9.01, which is a basic pH.
- We can arrive at this answer through the following calculation:
[tex]K_{a} +K_{b}= K_{c} \\K_{a} = 3.9*10^{-6}[/tex]
- From this we can calculate the values of K:
[tex]K_{b}= \frac{1*10^-14}{K_a} \\K_b= \frac{1*10^-14}{3.9*10^-6} \\K_b= 2.6*10^-9[/tex]
- We must also calculate the [tex]K_b[/tex] value related to [tex]HP^-[/tex] and [tex]OH^-[/tex]:
[tex]K_b=\frac{(HP^-)(OH^-)}{P^2-} \\2.6*10^-9=\frac{x*x}{0.042} \\(OH^-)=(HP^-)=1.04*10^-5\\[/tex]
- Now we can calculate the pOH:
[tex]pOH=-log[OH^-]\\pOH=-log[1.04*10^-5]\\pOH=4.99[/tex]
- Now we can calculate the pH:
[tex]pH+pOH=14\\pH+4.99=14\\pH=14-4.99\\pH= 9.1[/tex]
More information:
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