the pH of a .050 M ammonium chloride solution was experimentally determined to be 5.45. Write the equation for the chemical reaction that determines the pH

Answer 1

...............NH4^+ + H2O ==> NH3 + H3O^+

Answer 2

Thank yo, that' what I had, but I just wanted to make sure. Now how do I find the equilibrium constant? I have that the Ka = 3.55 *10^-6 and the ice table is as follows
NH4^+ + H2O ==> NH3 + H3O^+
I .05 0 0
C -x +x +x
E .05 - x x x

K = x^2 / .05 since x is small

Answer 3

x should be under each product not grouped together, this computer grouped them all together erroneously

Answer 4

You can use .......... as a way of spacing. It isn't perfect but it's better than letting the computer do it. It looks like this.
..................NH4^+ + H2O ==> NH3 + H3O^+
I................0.05..........................0...........0
C................-x.............................x...........x
E...............0.05-x........................x...........x

Keq= [(x)(x)/(0.05-x)]
Often these problems ask for Ka for the acid (the NH4^+) and that becomes
Ka for NH4^+ = (Kw/Kb for NH3) = x^2/0.05 which gives you Kb and that translates to Ka for the salt.

Answer 5

@DrBob222 okay, I have the ice table correct, i calculated the Ka for the NH4 which is 3.55*10^-6 and calculated the Kb for NH3 which is 2.82*10^-9. I'm just really stuck on finding the Equilibrium constant for the entire reaction.

Answer 6

I wrote Keq in my earlier post. Isn't Keq the same as Ka for NH4^+? Some may want to include (H2O) as a part of Keq; however, that's essentially a constant at 55.5 M and in most cases is not included as part of K.
Keq = (NH3)(H3O^+)/(NH4^+) = [(x)(x)/0.05-x]