Question

What is the ph of a 0.40 m ammonia solution?

Answer 1

Kb = 10^-4.74 = 1.8 x 10^-5 

NH3 + H2O <=> NH4+ + OH- 

1.8 x 10^-5 =x^2/ 0.40-x 
x = [OH-]= 0.00268 M 
pOH = 2.57 
pH = 14 - 2.57=11.4

Answer 2

pH of a 0.40 m ammonia solution is 11.43.

Ammonia is a weak base and has a $K_{b}=1.75 \times 10^{-5}$.

$\mathrm{K}_{\mathrm{b}}$ is the ionisation constant of ammonia.

We are given the amount of ammonia solution = 0.40 m

pH is basically the scale of acidity and tells the nature of a solution whether acidic or alkaline.

We calculate the pH of any solution by using the formula:

pH = 14 - pOH

Now, we calculate the pH using the formula

$[O H]^{-}=\sqrt{K_{b} \times c}$     where c is the concentration of the solution which is given to be 0.40m

$\begin{array}{l}{[O H]^{-1}=\sqrt{1.75 \times 10^{-5} \times 0.40}=0.00268} \\ {p O H=-\log \left(O H^{-}\right)=-\log (0.00268)=2.57} \\ {p H=14-p O H=14-2.57=11.43}\end{array}$.