Calculating pH: I have no idea where to start :(

I plan to add ammonium nitrate to my lawn. I add 18/00 grams of ammonium nitrate to a solution that has a final volume of 10.0L. What is the pH of this solution?

I have no idea where to start :(

plan to add ammonium nitrate to my lawn. I add 18/00 grams of ammonium nitrate to a solution that has a final volume of 10.0L. What is the pH of this solution?

http://en.wikipedia.org/wiki/Ammonium_hy…
Kb = [NH4+][OH-]/[NH3] = 1.8×10−5

Ammonium nitrate will produce NH4+1 ions when dissolved in water. The water will attract an H+1 off the NH4+1 ion to produce NH3 + H3O+1. Since the equilibrium constant for Ammonium hydroxide is very low, equilibrium will established.
NH4+1 + H2O → NH3 + H3O+1

NH4+ ↔ NH3 + H3O+1

Concentration of NH4NO3 = (mass ÷ molar mass) ÷ liters of solution
Molar mass = 14 + 4 + 14 + 48 = 80 grams
Initial concentration of NH4NO3 = (mass ÷ 80) ÷ 10 = [mass ÷ 800]

Initial concentration of NH4+1 = [mass ÷ 800]

When (mass ÷ 800) moles per liter of NH4 +1 ions is dissolved in H2O, x moles of NH3 and x moles of H3O+1 ion are produced and (mass ÷ 800) – x moles of NH4+1 ions .
Kb = [NH3] * [H3O+1] ÷ [NH4+1]

at equilibrium
[NH4+1] = (mass ÷ 800) – x
[NH3] = x
[H3O+1] = x

http://en.wikipedia.org/wiki/Ammonium_hy…
Kb = [NH4+][OH-]/[NH3] = 1.8×10−5
Ka * Kb = 1 * 10^-14
Ka = 1 * 10^-14 ÷ 1.8×10−5 = 5.56 * 10^-10

Ka = x * x ÷ [(mass ÷ 800) – x] = 5.56 * 10^-10

x^2 = [(mass ÷ 800) – x] * 5.56 * 10^-10
Since Kb is very small we can delete the – x
x^2 = (mass ÷ 800) * 5.56 * 10^-10

x = square root of (mass ÷ 800) * 5.56 * 10^-10
pH = -1 * log of x


. I add 18/00 grams of ammonium nitrate
I do not what the mass is!!

Never mind! it's 10 L, you did it all in one step! :-)
Thank you for your help :-)

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