How many atoms are in 1.14 mol of sulfur trioxide
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Answer:- The Ka for the acid is .
Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
Now, we make the ice table for this equation as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
I 0.25 0 0
C -X +X +X
E (0.25 - X) X X
where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.
X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.
Where, Ka is the acid ionization constant. Let's plug in the values.
Let's calculate the value of X first using the equation:
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on taking antilog ob above equation we get:
= 0.00195
So, X = 0.001195
Let's plug in this value of X in the equation:-
So, the value of Ka for butyric acid is .