Calculate the pH of a 0.44 M aqueous solution
of NH4Cl. Kb = 1.8 × 10−5
.
I've tried used the Ka=Kw/Kb equation then used the resulting 5.5e-10 in Kw=([H+][Cl-])/[NH4Cl] , using that to get the [H+], but it isn't correct and I don't know what else to do.
Chemistry AP - DrBob222, Thursday, February 5, 2015 at 10:34pm
You're right that Ka = Kw/Kb = 5.55E-10
But your equation is wrong.
Here is the hydrolysis equation which you need.
.........NH4^+ + H2O ==> NH3 + H3O^+
I........0.44.............0......0
C.........-x..............x......x
E.......0.44-x............x.......x
Then Ka = (x)(x)/(0.44-x)
Solve for x = (H3O^+) and convert to pH. Most NH4Cl solution are somewhere around pH 5 so you should get close to that (but probably not 5.0).
I can't figure out how to get the 0.44-x, which is the main issue, technically regardless of the wrong equation compared to yours, it /should/ come out to the same number. I cannot eliminate the -x using 10^3 rule because I am using the UT Quest system for online HW and it goes out to 6 decimal places for the final answer.
5.55E-10 = (x)^2/0.44-x
x is small in comparison to 0.44 so ignore it. Then x^2 = 5.55E-10*0.44
Then x^2 = 2.44E-10
Solve for x and convert to pH.
I don't know about a 10^3 rule but I know the x is small and can be ignored. Continuing with my work, then
x = sqrt(2.44E-10)
x = 1.56E-5 so you see
0.44-1.56E-5 = 0.43998 which is essentially 0.44 and this shows the x can be eliminated.
Convert x to pH.
Frankly, I think you've worried about something that isn't a problem. Or perhaps I've misunderstood what your problem is.
I made a silly mistake and inputted the correct answer of 4.805909915 as 4,805909915, I noticed as I looked at my previous answers, sorry for all of this.
I've figured out my mistake, mentioned above your post, mistypes when inputting answers are a terrible thing :p
I forgot to mention that there is another way to do it if yu want to go to the trouble. You can not ignore the x and make a quadratic equation out of it, then solve the quadratic. However, you will get the same answer.
5.55E-10 = x^2/(0.44-x)
5.55E-10*(0.44-x) = x^2
5.55E-10*0.44 - 5.55E-10*x = x^2
2.44E-10 -5.55E-10x = x^2
Rearrange to
x^2 + 5.55E-10x - 2.44E-10 = 0 and solve the quadratic. I used an on-line quadratic solver and obtained
x = 1.56E-5 = (H3O^+)
See no difference between ignoring the x and not ignoring the x.