25 cm3 of phosphoric acid, H3PO4, neutralised 42.6 cm3 of 0.2 M sodium hydroxide, NaOH. What was the concentration of the acid?

Equation for the reaction:
H₃PO₄ + 3NaOH → Na₃PO₄ + H₂O
Mole ratio H₃PO₄ : NaOH = 1 : 3

Moles of NaOH = (0.2 mol/dm³) × (42.6/1000 dm³) = 0.00852mmol
Moles of H₃PO₄ = (0.00852 mol) × (1/3) = 0.00284 mol
Concentration of H₃PO₄ = (0.00284 mol) / (25/1000 dm³) = 0.114 M

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OR:

(0.2 mol NaOH / 1000 cm³ NaOH solution) × (42.6 cm³ NaOH solution) × (1 mol H₃PO₄ / 3 mol NaOH) / (25/1000 dm³ H₃PO₄ solution)
= 0.114 M H₃PO₄

the reaction is
.. 1 H3PO4 + 3 NaOH --> 1 Na3PO4 + 3 H2O

then via dimensional analysis
.. 42.6mL NaOH... 0.2 mmol NaOH.. ..1 mmol H3PO4
 ---- ----- ---- ----- x ---- ---- ---- ---- ---- x ----- ----- ---- ---- = 0.114M H3PO4
.. 25mL H3PO4.. ... .1 mL NaOH..... . . 3 mmol NaOH

you calculate that by entering
.. 42.6 / 25 * 0.2 / 3 = 
and rounding to the appropriate sig figs..