QuoteIt means it can not be para so only two options are left - O-Bromotoluene or p-Bromotoluene. Analyze splitting in between 7 to 8 to identify it.

You mean o- bromotoluene and m-bromo- toluene, Do You?

MS spectra - Two isotopic molar ion peak are of same abundance so it means Br atom is there. Next abundant peak is at 91 which is 170-91 = 80
This also confirms Br is present.
H-1 NMR - Confirmed aromatic ring
IR also confirms - Aromatic ring
If we subtract mass of C6H4 and Br  from 170 we are left with(15) a methyl group.
Methyl on aromatic ring is confirmed by H-NMR also (signals in the range of 7ppm to 8ppm.
Two H are more shielded and two are more deshielded and all 4 Hs are different.So it means aromatic ring is either ortho or meta.
Signals of H1 NMR are indicates multiple signals due to multiple splitting.
It means it can not be para so only two options are left - O-Bromotoluene or p-Bromotoluene. Analyze splitting in between 7 to 8 to identify it.

Given this IR, NMR, and MS, what is my unknown molecule?

Ammonia is a base and concentrations of ammonia is important for pH calculations. Here pH is determined by calculation of [OH-] by making rice table and using Kb value.
K_b = [OH^-][NH₄⁺]/[NH₃]
[OH^-]= K_b[NH₃/[NH₄⁺]
so as  decreases  also decreases and pOH increases and hence pH decreases.
So B is the right option.

Definitely it is not a buffer solution as you are dealing with strong acid and strong base .
But C is not the Handerson equation
It is based on pH + pOH = 14 equation written in a different format.
Write reaction first and as you are mixing volume of two solutions so their concentrations are changing. Work with moles or  millimoles to know how much of the reactants have reacted and how much of the product is formed.
Remember – molarity = mmol /V in mL  =  mol / Vol (L)     
Means mol = M* V in L

Reaction
               NaOH(aq) +  HCl(aq) ->   Na+ (aq)  + Cl-(aq)  + H2O (l)

Initial (mMol)   11.0     10.0           0            0             -
Final (mMol))   1.0     0.00           10             10             -
HCl is the limiting reactant and finally solution has NaOH  (aq) left which is very strong base and it will control the pH .
Cl-(aq) and Na+ (aq) is a spectator here because they are weak conjugate base/acid  of strong acid and strong base. Water is also weak in front of NaOH.
Since volume has changed  you can calculate new concentrations. New volume is 21 mL = 0.021L
[ NaOH  ]  = mmol  /V mL  = [OH-]= 1.0 mmol/21mL 
Which can be written in terms of moles
[ NaOH  ]  = mol  /V L  = [OH-]= 0.0010 mol/0.021L 

pOH = -log[OH-] = -log (0.0010/0.021)
pH = 14- pOH
pH = 14 – (-log[OH-])
pH = 14+ log[OH-]
=14+ log (0.0010/0.021)
That's how C is the answer.

First write the reaction of HCl (acid) and ammonia (base)
NH3 (aq) + HCl(aq)    NH4+ (aq) + Cl-(aq) 
Now you are mixing volume of two solutions so their concentrations are changing. Work with moles and millimoles to know how much of the reactants have reacted and how much of the product is formed.
Remember – molarity = mmol /V in mL  =  mol / Vol (L)     
Means mol = M* V in L
Here volume is not given so let us assume the volume is  1L
Reaction
              NH₃(aq)+HCl(aq)-> NH₄⁺(aq)+Cl^-(aq)

Initial (Mol      0.12          0.10              0                    0
Final (Mol))     0.02          0.00            0.10                   0.10
HCl is the limiting reactant and finally solution has NH₃ (aq), NH₄⁺ (aq)  and Cl^-(aq) 
Cl^-(aq)   is a spectator here because it is a weak conjugate base of strong acid HCl.
pH is controlled by NH₃ (aq) and  NH₄⁺ (aq)  .However this is a base and its conjugate acid so it is a buffer and you can use Henderson - Hasselbalch equation to get the answer.
pH = pKa + log ([ NH₃ ]/[NH₄⁺ ] )
Since volume is doubled you can calculate new concentrations. New volume is 2L
[ NH3 ]  = 0.02 /2 = 0.01 M
[NH4+ ]  = 0.10 / 2 = 0.05 M
pH = 9.25 + log ( 0.01/ 0.05)
     Or you can simply take the ratio of moles since volume finally is same in the buffer solution for both acid and its conjugate base.
 pH = pKa + log (moles  NH₃ /moles NH₄⁺  )
          = 9.25 + log ( 0.02/ 0.10)
log ( 0.01/ 0.05)  = log ( 0.02/ 0.10)

         

The answer for this question is C but why would we use the Henderson Hasselbachs equation if it is not a buffer

I do not understand how to do this question

How do I do this question step by step?

You need to pick up the true statement. With reaction D statement given is not true. Only reaction A statement give is true and rest all are incorrect explanations are not true.