Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250ºC: NH4HS(s) H2S(g) + NH3(g) If 55.0 g of NH4HS(s) is placed in a sealed 5.0-L container, what is the partial pressure of NH3(g) at equilibrium

0.33 atm

Explanation:

Equation for the reaction is expressed as follows:

     ⇄    

= 0.11

The equilibrium constant for the reaction can be represented as:

=

= 1   since it is solid and solid and it is known that any solid has uniformity at equilibrium.

So; =

At equilibrium ;

Let x =

So;

=

=

atm

Therefore,  the partial pressure of NH3(g) at equilibrium = 0.33 atm

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.