Respuesta :

superman1987
First, we need to get moles of NaOH:

when moles NaOH = volume * molarity 

                                  = 0.02573L * 0.11 M

                                 = 0.0028 moles 

from the reaction equation:

H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)

we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH

 ∴ X mol H3PO4 reacts with → 0.0028 moles NaOH

∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol

now we can get the concentration of H3PO4:

∴[H3PO4] = moles H2PO4 / volume

               = 9.4 x 10^-4 / 0.034 L

               = 0.028 M
ardni313

The concentration of the unknown H₃PO₄ solution : 0.0344 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.

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. Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

We complete the possible questions

A 34.00 mL sample of an unknown H₃PO₄ solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 26.98 mL of NaOH solution is added.

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

H₃PO₄ -> 3H⁺ + PO₄⁻

NaOH is a strong base with valence 1 because it releases 1 OH- ions

NaOH ---> Na⁺ + OH-

We input the numbers into the formula:

M1.V1 n1= M2.V2 .n2

1 = H₃PO₄

2 = NaOH

M1.34 ml .3 = 0.130. 26.98 ml.1

M1 = 0.0344

So the molarity of the H₃PO₄ solution = 0.0344 M

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