Question
In a acidic solution with a PH of 3.4,78% of the acid is ionised.what is the pka of the acid?
Answers
GPT 3.5
To find the pKa of the acid, we first need to determine the concentration of the acid that is ionized and un-ionized in the solution.
Let x be the concentration of the acid that is ionized, and 0.78x be the concentration of the acid that is un-ionized.
Since 78% of the acid is ionized, we know that x/(x+0.78x) = 0.78
=> x/1.78x = 0.78
=> x = 0.78*1.78x
=> x = 1.39x
=> x = 0.58
Therefore, the concentration of ionized acid is 0.58 and the concentration of un-ionized acid is 1.00 - 0.58 = 0.42.
The pKa of the acid can be calculated using the formula:
pKa = -log(Ka)
Ka = [ionized acid][H+]/[un-ionized acid]
Ka = (0.58)(H+)/(0.42)
Since we know that the pH of the solution is 3.4 and the concentration of H+ ions is 10^(-3.4) = 3.98 x 10^(-4) M, we can substitute these values into the equation to find the pKa:
Ka = (0.58)(3.98 x 10^(-4)) / (0.42)
Ka = 1.74 x 10^(-4)
pKa = -log(1.74 x 10^(-4))
pKa = 3.76
Therefore, the pKa of the acid is 3.76.
Let x be the concentration of the acid that is ionized, and 0.78x be the concentration of the acid that is un-ionized.
Since 78% of the acid is ionized, we know that x/(x+0.78x) = 0.78
=> x/1.78x = 0.78
=> x = 0.78*1.78x
=> x = 1.39x
=> x = 0.58
Therefore, the concentration of ionized acid is 0.58 and the concentration of un-ionized acid is 1.00 - 0.58 = 0.42.
The pKa of the acid can be calculated using the formula:
pKa = -log(Ka)
Ka = [ionized acid][H+]/[un-ionized acid]
Ka = (0.58)(H+)/(0.42)
Since we know that the pH of the solution is 3.4 and the concentration of H+ ions is 10^(-3.4) = 3.98 x 10^(-4) M, we can substitute these values into the equation to find the pKa:
Ka = (0.58)(3.98 x 10^(-4)) / (0.42)
Ka = 1.74 x 10^(-4)
pKa = -log(1.74 x 10^(-4))
pKa = 3.76
Therefore, the pKa of the acid is 3.76.