Answer to Question #230247 in General Chemistry for Hasen

Question #230247
If a 0.2M NH4Cl solution has a ph of 5, what is the Ka of NH4?
1
Expert's answer
2021-08-28T06:26:43-0400

NH4Cl dissociates as:

"NH4Cl <--> NH4+ + Cl-"

From here:

"Ka =\\frac{ ([NH4^+][Cl^-])}{[NH4Cl]} = \\frac{[NH4^+]2}{[NH4Cl]}"


As pH = -log[H+]:

[H+] = 10-pH


From here, the concentration of [NH4+] in the solution equals:

[H+] = 10-5 M

As [H+] = [NH4+]= 10-5 M, [NH4Cl] = 0.200 M,

Ka of NH4+ equals:

"Ka =\\frac{ [NH4+]2}{[NH4Cl] }=\\frac{ (10-5)2 }{ 0.2} = 5 \u00d7 10^{-10}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS