What is the pH of aqueous solution of KC6H7O2 at 373 molar concentration?
A figure for sorbic acid C6H7O2H
pKa = 4.77
Google it, because it gives you several Pdfs. These are very useful most of the time, o I trust them. Doesn't your lesson explain this?
I mean doesn't your lesson explain pH of KC6H7O2 at 373 molar concentration?
373M would be quite a slurry. That is about 57kg per liter.
To determine the pH of an aqueous solution of KC6H7O2 at 373 molar concentration, we need to consider the nature of KC6H7O2 as a weak acid. In this case, KC6H7O2 dissociates in water to form the corresponding conjugate base C6H7O2- and a hydrogen ion (H+).
The dissociation equilibrium can be represented as follows:
KC6H7O2 ⇌ K+ + C6H7O2-
The equilibrium constant for this reaction is given by the acid dissociation constant, Ka:
Ka = [K+][C6H7O2-] / [KC6H7O2]
The pKa value is defined as the negative logarithm (base 10) of the Ka:
pKa = -log10(Ka)
In this case, the pKa for sorbic acid (C6H7O2H) is given as 4.77. Sorbic acid and KC6H7O2 are related, with KC6H7O2 being the potassium salt of sorbic acid. Therefore, we can assume that the pKa value for KC6H7O2 is also 4.77.
Now, using the Henderson-Hasselbalch equation, we can determine the pH of the solution based on the concentration of the acid and its conjugate base:
pH = pKa + log10([C6H7O2-] / [KC6H7O2])
At 373 molar concentration, we can substitute the values into the equation to calculate the pH.