What volume of 1.00M sodium hydroxide solution is needed to titrate 10.0 mL of 0.143 M H3PO4 solution?

Method 1 :

Number of moles of H₃PO₄ reacted = (0.143 mol/L) × (10.0/1000 L) = 0.00143 mol

The equation for the reaction :
3 NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O
Mole ratio NaOH : H₃PO₄ = 3 : 1
Then, number of moles of NaOH needed = (0.00143 mol) × 3 = 0.00429 mol

Volume of NaOH needed = (0.00429 mol) / (1.00 mol/L) = 0.00429 L = 4.29 mL


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Method 2 :

Number of milli-moles of H₃PO₄ reacted = (0.143 mmol/mL) × (10.0 mL) = 1.43 mmol

The equation for the reaction :
3 NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O
Mole ratio NaOH : H₃PO₄ = 3 : 1
Then, number of milli-moles of NaOH needed = (1.43 mmol) × 3 = 4.29 mmol

Volume of NaOH needed = (4.29 mmol) / (1.00 mmol/mL) = 4.29 mL


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Method 3 :

Volume of NaOH needed
= 10.0 mL × (0.143 mol H₃PO₄/ 1000 mL) × (3 mol NaOH / 1 mol H₃PO₄) × (1000 mL NaOH / 1.00 mol NaOH)
= 4.29 mL

n(acid) = 1.43 mmol
so
n(base) = 4.29 mmol
V(base) = 4.29 ml