Another 54 mL.
Explanation
According to the question, the number of molecules in each beaker indicates the relative concentration between each species at that scene.
Phosphoric acid shall lose hydrogen gradually as the titration proceeds. The order C → B → D → A thus appears to be correct.
The titration started with a solution of [tex]\text{H}_3\text{PO}_4[/tex] in scene C.
[tex][\text{H}_2\text{PO}_4^{-}] = [\text{H}_3\text{PO}_4] = 1/2 \; [\text{H}_3\text{PO}_4]_\text{initial} [/tex] in scene B. The 18.00 mL NaOH has neutralized exactly one-half of all the [tex]\text{H}_3\text{PO}_4[/tex] in the initial solution.
Assuming that there are 1 mol [tex]\text{H}_3\text{PO}_4[/tex]
It would take 0.5 moles of hydroxide ion to remove 0.5 moles of protons from 0.5 moles of [tex]\text{H}_3\text{PO}_4[/tex], such that the solution contains 0.5 mol of both [tex]\text{H}_2\text{PO}_4^{-}[/tex] and [tex]\text{H}_3\text{PO}_4[/tex]
The last scene contains only [tex]\text{HPO}_4^{2-}[/tex]. Converting 1 mol [tex]\text{H}_3\text{PO}_4[/tex] to [tex]\text{HPO}_4^{2-}[/tex] removes 2 mol protons and would consume 2 mol NaOH.
The question is asking for the amount of NaOH required for converting B to A. 0.5 mol of NaOH has already been added. Another 1.5 mol is required, which is three times the volume of NaOH required for converting C to B.
[tex]((2-0.5) / 0.5) \times 18 = 3 \times 18 = 54 \; \text{mL}[/tex]
