Answer:
The concentration c is equal to Ka
Explanation:
The acid will ionize as observed in the following reaction:
HA = H+ + A-
H+ is the proton of the acid and A- is the conjugate base
. The equation to calculate the Ka is as follows:
Ka = ([H+]*[A -])/[HA]
Initially we have to:
[H+] = 0
[A-] = 0
[HA] = c
During the change we have:
[H+] = +x
[A-] = +x
[HA] = -x
During balance we have:
[H+] = 0 + x
[A-] = 0 + x
[HA] = c - x
Substituting the Ka equation we have:
Ka = ([H+]*[A-])/[HA]
Ka = (x * x)/(c-x)
x^2 + Kax - (c * Ka) = 0
We must find c, having as [H+] = 1/2c. Replacing we have:
(1/2c)^2 + (Ka * 1/2 * c) - (c * Ka) = 0
(c^2)/2 + Ka(c / 2 - c) = 0
(c^2)/2 + (-Ka * c/2) = 0
c^2 -(c*Ka) = 0
c-Ka = 0
Ka = c
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Answer:
unbalanced I know you should know to
Because the density are v/mass not mass only so the one that floats have a density less than 1 and the one that sinks have a density more than 1 {(1) is the water density}
