NH4Cl is a salt. It hydrolyzes, (reacts with water)(actually, the NH4^+ hydrolyzes) to give
NH4^+ + HOH ==> NH3 + H3O^+
Then Ka, since the NH4^+ is acting as an acid = (NH3)(H3O^+)/(NH4^+)
(NH3) = y
(H3O^+) = y
(NH4^+) = 0.123-y
solve for y.
You don't know Ka but you can calculate it since KaKb=Kw.
what is the pH of an aqueous solution that is 0.123M NH4Cl
3 answers
my y = 2.25*10^-5
n what do i do
n what do i do
where did you get m?
[(NH3)(H3O^+)/(NH4^+)] = Kw/kb
(y*y/0.123-y)= 1 x 10^-14/1.8 x 10^-5
I don't know your book value for Kb for NH3 but I'm using 1.8 x 10^-5.
Then operate on the right side only to get
1 x 10^-14/1.8 x 10^-5 = 5.56 x 10^-10, then
[y^2/(0.123-y)] = 5.56 x 10^-10
Since y is small compared to 0.123, we can simplify by making 0.123-y = 0.123, then (or you can solve the quadratic if you wish)
(y^2/0.123)=5.56 x 10^-10
y^2 =5.56 x 10^-10*0.123 = 6.83 x 10^-11 and y = sqrt (...) = 8.27 x 10^-6
Convert that to pH which is approximately 5.10.
[(NH3)(H3O^+)/(NH4^+)] = Kw/kb
(y*y/0.123-y)= 1 x 10^-14/1.8 x 10^-5
I don't know your book value for Kb for NH3 but I'm using 1.8 x 10^-5.
Then operate on the right side only to get
1 x 10^-14/1.8 x 10^-5 = 5.56 x 10^-10, then
[y^2/(0.123-y)] = 5.56 x 10^-10
Since y is small compared to 0.123, we can simplify by making 0.123-y = 0.123, then (or you can solve the quadratic if you wish)
(y^2/0.123)=5.56 x 10^-10
y^2 =5.56 x 10^-10*0.123 = 6.83 x 10^-11 and y = sqrt (...) = 8.27 x 10^-6
Convert that to pH which is approximately 5.10.
7 answers