A0.10 m solution of a weak monoprotic acid has a ph of 3.40 at 25°c. what is the acid-ionization
constant, ka, for this acid?
a) 1.6 x 10-6
b) 4.0 x 10-4
c) 3.4 x 10-5
d) 1.2 x 10-3
e) 1.8 x 10-7
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Ответ:
The correct answer is A) 1.6 x 10-6
Explanation:
A weak monoprotic acid has the following dissociation equilibrium. At the beggining (t=0), the concentration of the monoprotic acid (HA) is equal to 0.10 M and the concentration of the ions H⁺ and A⁻ is zero (no dissociation). At a time t, dissociation occur and there is x concentration of H⁺ and A⁻ which is given by the dissociation constant Ka.
HA(aq) ⇄ H⁺(aq) + A⁻(aq)
t=0 0.10 M 0 0
t -x +x +x
eq 0.10 M-x x x
Ka=
As the pH is 3.40, we can calculate the concentration of both H⁺ and A⁻, as follows:
pH= - log (conc H⁺)= -log x
⇒ x = = 3.98 x 10⁻⁴
Now we introduce x in the previous equation to calculate Ka:
Ka=
Ka= 1.59 x 10⁻⁶ ≅ 1.60 x 10⁻⁶
Ответ:
29.9 moles
2C²H¹⁰ needs 13 moles of O²
4.6 C⁴H¹⁰ needs X moles of O²
X= 13× 4.6 ÷ 2 = 59.8 ÷ 2 = 29.9 moles