# 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