I think because its the only one to be liquid at normal temperatures.
Physical,physical, chemical, chemical, physical, chemical
There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But 
, so
and
STP is 0 °C and 1 bar, so
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
