NH3(aq)+H2O(l)⇄NH4+(aq)+OH−(aq)

Question

Kb=1.8×10−5
at 25C
Initial [NH3] | Equilibrium [NH4+] | Equilibrium [OH−] | pOH
0.15 | 1.6×10−3 | 1.6×10−3 | 2.78
0.30 | 2.3×10−3 | ? | ?

NH3 is a weak base that reacts with water according to the chemical equilibrium represented above. The table provides some information for two NH3(aq) solutions of different concentration at 25°C . Which of the following is true about the more concentrated 0.30MNH3(aq) , and why?

A. [OH−]=3.2×10−3M and pOH<2.78
B. [OH−]=3.2×10−3M and pOH>2.78
C. [OH−]=2.3×10−3M and pOH<2.78
D. [OH−]=2.3×10−3M and pOH<2.78

Why?
A. because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .
B. because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .

Answer 1

NH3(aq)+H2O(l)⇄NH4+(aq)+OH−(aq)..........Kb=1.8×10−5

Your table is hard to read. I think I can help that.
................. [NH3]............... [NH4+] ........[OH−]........ pOH
Initial.........0.15 ................ 1.6×10−3 ... 1.6×10−3 ..... 2.78
Initial.........0.30................. 2.3×10−3 ......2.3x10^-3 .....2.63
I see that both c and d or why are the same AND I confused about what you want. I've calculated the question marks you have so you can make the decision for the answer.

Answer 2

A. OH−]=3.2×10−3M and pOH<2.78 , because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq) .
B. [OH−]=3.2×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).
C. [OH−]=2.3×10−3M and pOH<2.78, because a higher [OH−] corresponds to a lower pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).
D. [OH−]=2.3×10−3M and pOH>2.78, because a higher [OH−] corresponds to a higher pOH for 0.30MNH3(aq) compared to 0.15MNH3(aq).