Question

Suppose that 6.2 mL of 2.1 M KOH(aq) is transferred to a 250 mL volumetric flask and diluted to the mark. It was found that 36.9 mL of this diluted solution solution was needed to reach the stoichiometric point in a titration of 6.2 mL of a phosphoric acid solution according to the reaction 3 KOH(aq) + H3PO4(aq) → K3PO4(aq) + 3 H2O(ℓ) Calculate the molarity of the solution. Answer in units of M.

Answer 1

Answer:

0.928 M

Explanation:

The concentration of acid can be determined by using the volume used and the concentration and volume used of base.

We will use the law of equivalence of moles.

M₁V₁=M₂V₂

M₁ = concentration of base used

V₁ = volume of base used

M₂ = concentration of acid used =? (to be determined)

V₂ = volume of acid used

The initial concentration of KOH used is diluted so let us find the final concentration of KOH after dilution

initial moles = final moles

initial concentration X initial volume = final concentration X final volume

6.2 X 2.1 = 250 X final concentration

final concentration = 0.052 M = M₁

V₁ = 36.9 mL

V₂ = 6.2 mL

Here with each mole of phosphoric acid three moles of KOH are used.

Therefore

3 M₁V₁ = M₂V₂

M₂ = $\frac{V_{2}}{3M_{1}V_{1}}=0.928M$