Respuesta :
Answer : The mathematical relationships gives the pOH of pure water at [tex]50^oC[/tex] is, [tex]pOH=14-pH[/tex]
Explanation :
As we are given that the pH at [tex]50^oC[/tex] is, 6.6.
Now we have to determine the mathematical relationships gives the pOH of pure water at [tex]50^oC[/tex].
As we know that:
The ionization of water is:
[tex]H_2O\overset{K_w}\rightarrow H^++OH^-[/tex]
The expression for dissociation constant for water is:
[tex]K_w=[H^+][OH^-][/tex]
taking logarithm on both side, we get:
[tex]\log K_w=\log [H^+]+\log [OH^-][/tex]
Taking negative sign on both side, we get:
[tex]-\log K_w=-\log [H^+]+(-\log [OH^-])[/tex]
[tex]pK_w=pH+pOH[/tex]
As we know that the value of [tex]pK_w[/tex] is 14 at 25-50°C.
So,
[tex]14=pH+pOH[/tex]
or,
[tex]pOH=14-pH[/tex]
[tex]14=6.6+pOH[/tex]
[tex]pOH=7.4[/tex]
Therefore, the mathematical relationships gives the pOH of pure water at [tex]50^oC[/tex] is, [tex]pOH=14-pH[/tex]
Ionization is the process in which an atom or a molecule gains either a positive or negative charge.
The mathematical expression for the given pure water ionization at 50°C is:
pOH = 14 - pH
Given that:
pH at 50°C = 6.6
The ionization reaction for the pure water is:
H₂O [tex]\rightarrow[/tex] H⁺ + O⁻
The dissociation constant for the water can be given as:
K[tex]_\text w[/tex] = [ H⁺] [O⁻]
Taking log:
log K[tex]_\text w[/tex] = log [ H⁺] + log [O⁻]
Intorducing the negative sign:
- log K[tex]_\text w[/tex] = -log [ H⁺] + (-log [O⁻])
pK[tex]_\text w[/tex] = pH + pOH
Now, the value of pK[tex]_\text w[/tex] is 14 at 25-50°C, such that:
14 = pH + pOH
pOH = 14 - pH
14 = 6.6 + pOH
pOH = 7.4
Therefore, the pOH of water at 50°C is 7.4.
To know more about dissociation constant, refer to the following link:
https://brainly.com/question/4218061
