Respuesta :
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
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
[tex]\large{\boxed{\bold{Ma~Va~na~=~Mb~Vb~nb}}}[/tex]
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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