Calculate the volume (in mL) of 0.170 M NaOH that must be added to 413 mL of 0.0413 M (MOPS) to give the solution a pH of 7.55.?

Assume that y mL of NaOH is added.
Denote MOPS as HA.

Initial number of moles of HA = (0.0413 mol/L) × (413/1000 L) = 0.0171 mol
Number of moles of NaOH added = (0.170 mol/L) × (y/1000 L) = 0.000170y mol

When NaOH is added :
HA + NaOH → NaA + H₂O

After reaction :
Number of moles of HA in the solution = (0.0171 - 0.000170y) mol
Number of moles of A⁻ in the solution = 0.000170y mol
Hence, [HA] / [A⁻] = (0.0171 - 0.000170y) / 0.000170y

Consider the dissociation of HA :
HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq) ...... pKₐ = 7.18

pH = pKₐ - log([HA] / [A⁻])
7.55 = 7.18 - log([HA] / [A⁻])
log([HA] / [A⁻]) = -0.37
[HA] / [A⁻] = 10⁻⁰·³⁷
(0.0171 - 0.000170y) / 0.000170y = 10⁻⁰·³⁷
0.0171 - 0.000170y = 0.000170y × 10⁻⁰·³⁷
0.000170y × 10⁻⁰·³⁷ + 0.000170y = 0.0171
y = 0.0171 / (0.000170 × 10⁻⁰·³⁷ + 0.000170)
y = 70.5

Volume of 0.170 M NaOH added = 70.5 cm³