If the #"pH"# of an aqueous solution of ammonium chloride is #4.60#, what is the #K_h# value of #"NH"_4^(+)#?

Question

Peyton Adams · Accepted Answer

I assume it's an error in your notes? The relationship between #K_a#, #K_b#, and #K_w# is that #K_a*K_b = K_w = 10^(-14)#.

Even though you are not given #K_a#, you can still look it up. I happen to remember the #K_b# for #"NH"_3# though, so here's an opportunity to demonstrate the relationship.

The #K_b# for ammonia, #"NH"_3#, is related to the #K_a# for ammonium, #"NH"_4^(+)#.

#K_b = 1.8 xx 10^(-5)#

#K_w = 10^(-14)#

Consequently:

#K_a = K_w/K_b = 10^(-14)/(1.8 xx 10^(-5)) = 5.55 xx 10^(-10)#

So, the #K_a# for #"NH"_4^(+)# is #5.55 xx 10^(-10)#, and this is what you would expect for your answer.. You should also be able to look this up in the Appendix of your textbook.

Presume for a moment that you were unaware of that. After building your ICE table:

#"NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "H"_3"O"^(+)(aq)#

You ought to obtain:

#K_a = (x^2)/(0.1 - x) = ?#

If you aren't sure how I got here, tell me. #"NH"_4^(+)# is a weak acid since #"NH"_3# is a weak base. When you look at the #K_b# for a weak base, it is moderately small. Therefore, for the conjugate weak acid, it is even smaller.

When we calculate #K_a#, do NOT use the small #x# approximation, because it will not work unless #K_a# is small enough. Beyond that, you should only use it when you use #K_a# in the first place.

We know #"pH" = 4.60 = -log["H"^(+)]#. So:

#["H"^(+)] = x = 10^(-4.60)#

Consequently:

#color(blue)(K_a) ~~ (10^(-4.60))^2/(0.1 - 10^(-4.60)) = color(blue)(6.310 xx 10^(-9))#

Serenity Allen · Answer

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 #sqrt("Ka" xx "molarity") = m# #-log(m) = pH# so by solving for m when given the pH we can find the answer of square root.  so -log(x) =  4.60 x = #10^-4.60#
x = 0.00003 #sqrt("Ka" xx " 0.1M") =  0.00002511886# solve for Ka #Ka xx 0.1 =  0.00002511886^2# #6.30957344e-9 = Ka# #K_h = K_w/ (Kb)#
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

Olivia Anton · Answer

undefined The $ K_h $ value for $ NH_4^+ $ can be calculated using the formula:

$$ K_h = 10^{-\text{pH}} $$

Plugging in the pH value of the ammonium chloride solution:

$$ K_h = 10^{-4.60} $$

$$ K_h \approx 2.51 \times 10^{-5} $$