calculate the pH of 0.04M solution of NH4Cl if pk_b of NH3 is 4.74

Dear student
Please find the solution to the asked query:

NH₄⁺ +HOH $\to$   NH₃ + H₃O⁺
​We make an ICE table to find the concentration of OH- in solution:

	NH4+	NH3	H3O+
Initial	0.04	0	0
Change	-x	+x	+x
Equilibrium	0.04-x	x	x

$$\begin{gathered} K_b =\frac{\left[N{H}_3\right]\left[H_3{O}^{+}\right]}{\left[N{H_4}^{+}\right]}=\frac{x^2}{0.04-x} \\ [Given p{K}_b = 4.74, \\ p{K}_b = -\log (K_b) \\ {K}_b = {10}^{-p{K}_b} \\ ={10}^{-4.74} \\ =1.8\times {10}^{-5} ] \end{gathered}$$

Substituting the value of K_b in the equation above we get:
$$\begin{gathered} 1.8\times {10}^{-5}=\frac{x^2}{0.04-x} \\ [x in the denominator is negligible quantity, hence it can be omitted] \\ 0.72\times {10}^{-6} = x^2 \\ x = 0.85 \times {10}^{-3} \\ x =\left[O{H}^{-}\right] = 0.00085 \\ pOH =-\log \left[O{H}^{-}\right]=-\log [0.00085] \\ =3.07 \\ Thus pH =14-pOH \\ =14- 3.07 \\ =10.93 \end{gathered}$$


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