Answer is: because pH value of solution is changing.
Balanced chemical reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
pH of sodium hydroxide solution is above seven (basic), when solution of hydrochloric acid is added, pH slowly dropping until it became neutral solution (pH is equal seven) and that is endpoint of titration.
Divide the volume by the area. Using scientific makes things a bit cleaner.


Then

Now, 1 m = 10⁹ nm, so

Answer:
Rxn will shift toward reactant side of equation b/c Kc < Qc
Explanation: