Thw answer is PHj78 JJ CP30 R2D2
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
Answer: The concentration of
ions in vinegar is 0.001 M.
Explanation:
Given: pH = 3.0
pH is the negative logarithm of concentration of hydrogen ion.
The expression for pH is as follows.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the value into above expression as follows.
![pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C3.0%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%20antilog%20%28-%203.0%29%5C%5C%3D%200.001%20M)
Thus, we can conclude that the concentration of
ions in vinegar is 0.001 M.