If the #"pH"# of an aqueous solution of ammonium chloride is #4.60#, what is the #K_h# value of #"NH"_4^(+)#?
2 Answers
I assume it's an error in your notes? The relationship between
Even though you are not given
The
#K_b = 1.8 xx 10^(-5)#
#K_w = 10^(-14)#
Therefore:
#K_a = K_w/K_b = 10^(-14)/(1.8 xx 10^(-5)) = 5.55 xx 10^(-10)#
So, the
Now, pretend you didn't know that. Once you construct your ICE table:
#"NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "H"_3"O"^(+)(aq)#
You should get:
#K_a = (x^2)/(0.1 - x) = ?#
If you aren't sure how I got here, tell me.
When we calculate
We know
#["H"^(+)] = x = 10^(-4.60)#
Therefore:
#color(blue)(K_a) ~~ (10^(-4.60))^2/(0.1 - 10^(-4.60)) = color(blue)(6.310 xx 10^(-9))#
First calculate Ka then we can calculate Kh
I'm going to find the Ka from a different way you may dont know which is only applicable to weak acids
When we solve for Ka
we do this
so by solving for m when given the pH we can find the answer of square root.
so -log(x) = 4.60
x =
x = 0.00003
solve for Ka
and u will again get the Ka
For acids like this Ka = Kh
Oswald's law of dillution can be used to calculate Ka of weak electrolytes or weak acids like this one