Explanation:
The reaction equation will be as follows.

Calculate the amount of
dissolved as follows.

It is given that
= 0.032 M/atm and
=
atm.
Hence,
will be calculated as follows.
=
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
= 
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

Regards!
Answer:
Ka = ( [H₃O⁺] . [F⁻] ) / [HF]
Explanation:
HF is a weak acid which in water, keeps this equilibrium
HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq) Ka
2H₂O (l) ⇄ H₃O⁺ (l) + OH⁻ (aq) Kw
HF is the weak acid
F⁻ is the conjugate stron base
Let's make the expression for K
K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]
K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]
K . [H₂O] = Ka
Ka, the acid dissociation constant, includes Kwater.
<span>valence electrons is our answer cuz i am good at chemstry</span>