Step-by-step 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) = ([H^(+)]^(2))/([CO_(2)])](https://img.qammunity.org/2020/formulas/chemistry/college/30h27wdsy07ecxbxigmjhtnmvnyovroh4p.png)
=

=

Since, we know that pH =
![-log [H^(+)]](https://img.qammunity.org/2020/formulas/chemistry/high-school/gopcqwordys161r0cdra5oyonsnub5hpxq.png)
So, pH =

= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.