Determine the mass of Н3РО4 in the solution, if when titrating it with methyl orange there is 25.50 cm3 of a 0.2000 M NaOH solution

Decision:
M (H3PO4) = 98 g / mol.
The reaction equation has the form:
Н3РО4 + 3NaOH = Na3PO4 + 3H2O
From the equation it follows that titration of 1 mol of Н3РО4 requires 3 mol of NaOH, n (Н3РО4) = 1 / 3n (NaOH).
1) We find the amount of NaOH used for titration:
n (NaOH) = [CM (NaOH). V (NaOH)] / 1000 = (25.5. 0.2000) / 1000 = 0.0051 mol.
2) we find the amount of Н3РО4, we get:
n (H3PO4) = 1 / 3n (NaOH) = 0.0051 / 3 = 0.0017 mol.
3) We find the mass of Н3РО4 in solution, we get:
m (H3PO4) = n (H3PO4). M (H3PO4) = 0.0017. 98 = 0.1666 g.
The problem can be solved in one action, using one formula:
m (H3PO4) = V (p-pa) [NaOH]. CM (NaOH). M (Н3РО4) / (1000. 3) = (25.5. 0.2000. 98) / (1000. 3) = 0.1666 g.
Answer: m (Н3РО4) = 0.1666 g