Answer:
11.44
Explanation:
[tex]K_b[/tex] = Base ionization constant of ammonia = [tex]1.8\times 10^{-5}[/tex]
m = Mass of ammonia = 5 g
V = Volume of ammonia = 709 mL = [tex]709\times 10^{-3}\ \text{L}[/tex]
M = Molar mass of ammonia = 17.03 g/mol
Molaritiy of ammonia
[tex]C=\dfrac{n}{V}\\\Rightarrow C=\dfrac{\dfrac{m}{M}}{V}\\\Rightarrow C=\dfrac{\dfrac{5}{17.03}}{709\times 10^{-3}}\\\Rightarrow C=0.414[/tex]
We have the relation
[tex]pOH=\dfrac{1}{2}(pK_b-\log C)\\\Rightarrow pOH=\dfrac{1}{2}(-\log (1.8\times 10^{-5})-\log 0.414)\\\Rightarrow pOH=2.56[/tex]
[tex]pH+pOH=14\\\Rightarrow pH=14-pOH\\\Rightarrow pH=14-2.56\\\Rightarrow pH=11.44[/tex]
The pH of the solution is 11.44.