Calculate the pH of a 0.46 M solution of NH 4Cl. ( K b for NH 3 = 1.8 × 10 –5) a.
9.59
b.
4.41
c.
4.80
d.
0.34
e.
9.20

pH of the solution = 4.8

NH₄Cl is a salt of a weak base and a strong acid so that it will be partially hydrolyzed

Reaction:

NH₄Cl -----> NH₄⁺ + Cl⁻

NH₄ will be hydrolyzed in water

NH₄⁺ + H₂O ------> NH₃ + H₃O⁺ (H⁺)

This solution releases H⁺ ions so they are acidic

Formula

Kw = The water equilibrium constant  = 10⁻¹⁴

Kb = the ionization constant of a base = 1.8.10⁻⁵

M = 0.46

[H+]=√2.55 x 10⁻¹⁰ = 1.599 x 10⁻⁵

pH = - log [H+]

pH = 5-log 1.599 = 4.796≈ 4.8

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.

"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"