Determine the ph of a 0.62 m nh4no3 solution at 25°c. The kb for nh3 is 1.76 × 10-5.

ph = 4.73

Explanation:

Determine the pH of a 0.62 M NH4NO3 solution at 25C

The Kb for NH3 is 1.76 x 10^-5

NH4+ is the conjugate acid of the base NH3

Ka for NH4+ = K water / K NH3 conjugate base

Ka = 1 × 10⁻¹⁴ / 1.76 10⁻⁵

     = 5.68 × 10⁻¹⁰

which is he Ka for

NH4 + H2O --> NH4OH + H⁺

Ka = [NH4OH] [H+] / [NH4)+]

5.68 × 10⁻¹⁰ = [X] [X] / [0.62]

X² = 3.52 × 10⁻¹⁰

X = [H+] = 1.88   × 10⁻⁵

p of that H+ = 4.73

ph = 4.73

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !