A0.12 m solution of a monoprotic acid is 2.3% ionized. what is the ka of this acid?
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Ответ:
HA ↔ H+ + A-
Ka =
Now, initial conc. of HA = 0.12 M
Of these, 2.3 % got ionized.
∴ [H+] = [A-] = 0.12 X 0.023 = 0.00276 M
Thus [HA] left after ionization = 0.12 - 0.00276 = 0.11724 M
Now Ka =
Ответ:
It would be true
Explanation:
A scientific hypothesis is initial building block in a scientific method many describe it as an educated guess a hypothesis also include an explanation of why the guess may be correct .