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