Achieve Week 2 Question 8 Help Needed

[805751553]

Can someone explain to me how they solve this so I can figure out what I did wrong in my calculations? NH3 is a weak base ( Kb=1.8×10−5 ) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.019 M in NH4Cl at 25 °C?

[Julia_Wong_1H]

To solve this, start by setting up an ICE table to find equilibrium concentrations. Find Ka by finding Kw/Kb, or (1x10^-14)/(1.8×10^−5). Since Ka is equal to (x^2)/(0.019-x), you can set up the value you found from Ka = (x^2)/(0.019-x). Solve for x (you don't need to use quadratic equation in this case since x is so small). Take the -log of x and you will find the pH!

[Brandy_Lopez_1C]

Your ice table should look something like this

NH+4(aq) ↽−−⇀ H+(aq) + NH3(aq)
i 0.019 0 0
c −x +x +x
e. 0.019−x x x

And your setup for ka should look like this:
ka = [H+][NH3]/[NH4]= x^2/0.019-x --> x^2/0.019