Can someone help me with this problem?

[Patrick Essio 1H]

Hello!
I am doing one of the practice midterm worksheets and I came across this problem that I am struggling with. Can someone provide some insight?

[Patrick Essio 1H]

It is number 15 on this worksheet.

[Ananya Ravikumar 1I]

I am getting D. 0.074M
since pOH=2.94, [OH-]=10^(-2.94)=0.001148M
[NH4+]=[OH-] at equilibrium
The reaction shows the basic property of NH3, and we are given Ka, so we find Kb=10^-14/Ka=1.78*10^-5
Since the Kb is small, we can consider that the initial and equilibrium concentration of NH3 is the same. Let it be x.
Kb = [NH4+][OH-]/x
plug in the numbers and x will come to be 0.074M

[Zubaida Morataya Bashir 2H]

since you are given the pOH, you can get the concentration of OH from this by doing 10^-2.94. This would also give you the concentration of NH4+. It's important to note that they gave you Ka when you need Kb since this has a basic nature, so you'd need to convert Ka to Kb. Hope this helps!