The titration of 80.0 mL of an unknown concentration H3PO4 solution requires 126 mL of 0.218 M KOH solution. What is the concentration of the H3PO4 solution (in M)?

The concentration of the H₃PO₄ solution (in M) is 0.114 M

To determine the concentration of the H₃PO₄ solution,

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

Now, from the formula

Where is the concentration of acid

is the concentration of base

is the volume of acid

is the volume of base

 is the mole ratio of acid

is the mole ratio of base

From the question

From the balanced chemical equation for the reaction

Now, put all the above parameters into the equation

We get

Then,

Hence, the concentration of the H₃PO₄ solution (in M) is 0.114 M

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Ca = 0.1145M

Explanation:

Volume of Acid (Va) = 80.0ml

Concentration of acid (Ca) = ?

Volume of base (Vb) = 126ml

Concentration of base (Cb) = 0.218M

The balanced chemical equation is given as;

3KOH + H3PO4 --> K3PO4 + 3H2O

The formular relaing the parameters is given as;

(CaVa) / (CbVb) = Na/Nb

Where Na = Number of moles of Acid = 1

Nb = Number of moles of Base = 3

Making Ca subject of interest;

Ca = (Na * Cb * Vb) / (Nb * Va)

Ca = (1 * 0.218 * 126) / (3 * 80)

Ca = 27.468 / 240

Ca = 0.1145M

14 days is the half-life of phosphorus-32