Posted by **help!!!** on Monday, February 16, 2009 at 8:42pm.

what is the pH of an aqueous solution that is 0.123M NH4Cl

- chem -
**DrBob222**, Monday, February 16, 2009 at 9:04pm
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.

- chem -
**help!!!**, Monday, February 16, 2009 at 9:34pm
my y = 2.25*10^-5

n what do i do

- chem -
**DrBob222**, Monday, February 16, 2009 at 10:08pm
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.

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