Respuesta :
First, we need the reaction to start the exercise.
HCOOH(aq) + H2O(l) = H3O+(aq) + HCOO-(aq)
HCOOK => HCOO- + H+
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pH)
We have here an ionic equilibrium, HCOOH is a weak acid.
pKa = 3.77 (from tables)
The concentration of HCOOH 0.3 M
The concentration of HCOO- = 0.52 M HCOOK.
If we propose Ka for HCOOH(aq)
[tex]Ka\text{ = }\frac{\lbrack H_3O^+\rbrack x\lbrack HCOO-\rbrack^{_{}}}{\lbrack HCOOH\rbrack}[/tex]We clear (H3O+) and we apply -log and we get this:
[tex]pH\text{ = }pKa+log(\frac{[HCOO^-]}{[HCOOH]})[/tex][tex]pH=3.77+log(\frac{[0.52]}{[0.30]})\text{ }[/tex]Answer: pH = 4.01
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pOH)
pH + pOH = 14
(This formula appears from the ionic equilibrium of water)
pOH = 14 - 4.01 = 9.99
Answer: pOH= 9.99
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