A 25.00 ml sample of h3po4 solution is neutralized by exactly 54.93 ml of 0.04345 m ca (oh) 2. what is the molarity of the h3po4 solution?

The neutralization equation is:

3 Ca (OH) ₂ + 2 H₃PO₄ → Ca₃ (PO₄) ₂ + 6 H₂O

From this equation we can see that 3 moles of Ca (OH) ₂ react with 2 moles of H₃PO₄

Numbers of mmol of Ca (OH) ₂ = M x V = 0.04345 x 54.93 = 2.387 mmol

Number of mmol of H₃PO₄ = 2.387 x (2/3) = 1.591 mmol

Molarity of solution = n (mmol) / V (ml) = 1.591 / 25 = 0.0636 M