Calculate the pH of 130 ml of 0.34 M NH3 before and after the addition of 4.2 g of NH4NO3.

Question

Assume that the volume remains constant.

I *really* need to see how this is worked ...

To help, I've already figured the amu for NH4NO3 is 80.06

and 4.2 grams of that substance equals 0.05246 moles of NH4NO3

given the volume remains constant, I obtained a molarity for NH4NO3 by dividing the moles by the 0.130 L

0.05236 moles / 0.130 L = 0.4035 M NH4NO3

Answer 1

NH3 forms the bronsted base NH4 when it comes in contact with water. You can find the pH of the solution before the salt is added by taking subtracting the pOH from 14. The pOH is -log[base]. Therefore:

pH = 14 - -log(.34M) = 13.53

When you add the salt, you get the common ion effect. Make a chart showing the concentrations of the reactants and products like below

NH4NO3 ->  NH4  +  NO3-

.4035M       .34M       0M         initial

    -x              +x         +x        change

.4035 - x     .34 +x      x          equilibrium

This is how the concentrations of the chemicals change as the reaction occurs. Unfortunately I just got lost because I'm rusty, and I just realized that NH3 doesn't have a Kb value, but I hope this helps.