Asample 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?

The concentration of the H₃PO₄ solution is 0,245M

Explanation:

The first titration is:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

0,495g of KHP are:

0,495g×= 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!

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