Suppose you have 50.1 mL of a solution of H3PO4. You titrate it with a 0.10...

Suppose you have 50.1 mL of a solution of H₃PO₄. You titrate it with a 0.10 M solution of KOH and use 15.4 mL to reach the endpoint.

What is the concentration of H₃PO₄?

What was the pH of the original H₃PO₄ solution?

Expert Solution

balanced chemical equation :

H3PO4 + 3 KOH ----------------------> K3PO4 + 3H2O

C1 V1 / n1 = C2 V2 / n2

C1 x 50.1 / 1 = 0.10 x 15.4 / 3

C1 = 0.0102 M

concentration of H3PO4 = 0.0102 M

H3PO4 ---------------------> H3PO4- + H+

0.0102 0 0 --------------> initial

0.0102-x x x ----------------> equilibrium

Ka1 = x^2 / 0.0102 -x = 6.9 x 10^-3

x^2 + 6.9 x 10^-3 x - 7.07 x 10^-5 = 0

x = 5.64 x 10^-3

[H+] = 5.64 x 10^-3 M

pH = -log [H+]

pH = -log (5.64 x 10^-3)

pH = 2.25

pH of the original H₃PO₄ solution = 2.25

queen_honey_blossom answered 2 years ago

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Question

Suppose you have 50.1 mL of a solution of H3PO4. You titrate it with a 0.10...

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Expert Solution

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