I got the rest, I just need help on the last one. My first post is down on this page.
6th one:I don't know how to write the equation. It would be acid over base but which is the acid and which is the base?
My answer:
I got the Ka to be 2.7x10^-1.
The equation I got was:
C6 H 5 NH 2(aq) +H2O>>>CH 3 NH 2(aq)
Would this be right?
You appear to be trying to use the Henderson-Hasselbalch equation but that isn't necessary here so we don't need to distinguish between the acid and the base.
I don't get the same answer for Ka.
Kw = KaKb
Kw is 1 x 10^-14
Kb is 4.27 x 10^-4
The equation is C6H5NH2 + HOH ==> C6H5NH3^+ + OH^-
To determine which is the acid and which is the base in the equation, we can look at the reaction and identify any substances that release hydrogen ions (H+) when dissolved in water. These substances are acids. On the other hand, substances that release hydroxide ions (OH-) are bases.
In the given equation:
C6H5NH2(aq) + H2O(l) → C6H5NH3+(aq) + OH-(aq)
In this reaction, C6H5NH2 is acting as a base because it accepts a proton and forms C6H5NH3+ (phenylammonium ion). H2O, on the other hand, is acting as the acid because it donates a proton and forms OH- (hydroxide ion).
Now, regarding the value of Ka, we don't need the Henderson-Hasselbalch equation in this case. The Henderson-Hasselbalch equation is typically used for calculating the pH of a buffer solution, which doesn't apply here.
Instead, we can use the equilibrium constant expression for the dissociation of the base C6H5NH2:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
Since Kb can be related to Ka through the ion product of water (Kw), we can use the following relation:
Ka = Kw / Kb
Given that Kw = 1 x 10^-14 and Kb = 4.27 x 10^-4, we can calculate Ka:
Ka = (1 x 10^-14) / (4.27 x 10^-4)
After evaluating this expression, we find Ka to be approximately 2.34 x 10^-11.
Therefore, the equation you provided, C6H5NH2(aq) + H2O → C6H5NH3+(aq), is correct. Just keep in mind the acid-base distinction mentioned above and the different value of Ka.