Question

A sample of 0.495 grams of solid KHP is weighed into an Erlenmeyer flask. This sample is titrated with a sodium hydroxide solution, and 28.56 mL of NaOH are required to reach the endpoint. The sodium hydroxide solution is then used to titrate a sample of phosphoric acid of unknown concentration. It requires 29.88 mL of NaOH to react with 10.33 mL of H3PO4 solution. What is the concentration of the phosphoric acid?

Answer 1

The concentration of the H₃PO₄ solution is 0,245M

Step-by-step explanation:

The first titration is:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

0,495g of KHP are:

0,495g×
`(1mol)/(204,22g)`= 2,42x10⁻³ moles of KHP

As 1 mole of KHP reacts with 1 mole of NaOH, moles of NaOH are 2,42x10⁻³ moles.

As volume required was 28,56mL, the concentration of the NaOH solution is:

2,42x10⁻³ moles / 0,02856L = 0,0849M

The titration of the phosporic acid with NaOH occurs as follows:

H₃PO₄ + NaOH → H₂PO₄⁻ + Na⁺ + H₂O

If were required 29,88mL of NaOH, the moles of NaOH spent were:

0,0849M×0,02988L = 2,54x10⁻³ moles of NaOH that are the same than H₃PO₄ moles.

As the volume of the solution of H₃PO₄ was 10,33mL, the concentration of the H₃PO₄ solution is:

2,54x10⁻³ moles of H₃PO₄ / 0,01033L = 0,245M

I hope it helps!