A25.00 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

With an acid-base titration formula, The molarity of the H₃PO4₄ solution = 0.06372 M

Further explanation

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

An acid-base titration there will be a change in the pH of the solution. When the acid is alkaline the pH drops, and vice versa if the acid is alkaline then the pH will rise.

From this pH change a Titration Curve can be made which is a function of acid / base volume and pH of the solution

Acid-base titration formula

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

Known variable

A 25.00 ml sample of H₃PO₄ 54.93 ml of 0.04345 m Ca(OH)₂.

Asked

The molarity of the H₃PO₄ solution

Answer

H₃PO₄ is a weak acid with valence 3 because it releases 3 H⁺ atoms

H₃PO₄ -> 3H + + PO₄⁻

Ca(OH)₂ is a strong base with valence 2 because it releases 2OH⁻ ions

Ca(OH)₂ ---> Ca₂ + + 2OH-

We enter the numbers into the formula:

M₁.V₁.n₁= M₂.V2₂.n₂

1 = H₃PO4₄

2 = Ca(OH)₂

M1.25 ml .3 = 0.0435. 54.93 ml.2

M1 = 0.06372

So the molarity of the H₃PO4₄ solution = 0.06372 M

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