Consider a 0.25 M solution of H2CO3 (Kg = 4.4x10-7) What is the molarity of H?...

Question

Question

Answer 1

Answer #1

Given that the concentration of carbonic acid (H₂CO₃) C = 0.25 M

The dissociation constant of weak acid

K_a = 4.4 X 10^-7

[H⁺] = √(Ka C)

= √(4.4 X 10^-7 X C)

= √(44 X 10^-8 X 0.25)

= 3.3166 X 10^-4 M

1) Answer:

The concentration of [H⁺] = 3.3166 X 10^-4 M

P^H = - log (H⁺) = - log ( 3.3166 X 10^-4)

= 4 - log(3.3166) = 4 - 0.52069 = 3.479

2) Answer:

Conclusion: The P^H of the solution = 3.479

Answer 2

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