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

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

1 is the answer to this problem