Finding final pH of the buffer solution without applying Henderson equation A is beaker with $\pu{0.1 M}$ $\pu{25 cm^3}$ solution of $\ce{NH3}.$ B is beaker with $\pu{0.1 M}$ $\pu{5 cm^3}$ $\ce{HCl}.$ Now the solution of A and B is mixed. What will be the final $\mathrm{pH}$ of the mixture? $K_\mathrm{b} = \pu{3.3E-5}.$

I believe the mixture will become a buffer solution and we should use Henderson equation to solve this.
$$n(\ce{NH3}) = \pu{0.1 mol L^-1} × (\pu{25E-3 L}) = \pu{2.5E-3 mol}\tag{1}$$
$$n(\ce{HCl}) = \pu{0.1 mol L^-1} × (\pu{5E-3 L}) = \pu{5E-4 mol}\tag{2}$$
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log\frac{n(\ce{NH4Cl})}{n(\ce{NH4OH})}\tag{3}$$
$$\ce{NH3 + HCl -&gt; NH4Cl}$$
$$n(\ce{HCl}) = n(\ce{NH4Cl}) = \pu{5E-4 mol}\tag{4}$$
$$n(\ce{NH3}) = n(\ce{NH4OH}) = \pu{2.5E-3 mol} - \pu{5E-4 mol} = \pu{2E-3 mol}\tag{5}$$
$$
\begin{align}
\mathrm{pOH} &amp;= \mathrm{p}K_\mathrm{b} + \log\frac{\pu{5E-4 mol}}{\pu{2E-3 mol}}\\
  &amp;= 4.4819 - 0.602\\
  &amp;\approx 3.88 \tag{6}
\end{align}$$
$$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 3.88 = 10.12\tag{7}$$
However, the book from which I am solving this problem suggests that this problem should not be solved by Henderson equation without providing any reason. Can anyone tell me why this solution should not be considered a basic buffer? I would really appreciate some opinion on this.
P.S. The problem was given in Bangla. I tried my best to translate it. This is an admission question of KUET-2019 Bangladesh. My textbook has this question, but I'm not sure about the solution. Please note that this is an admission question of previous year (2019).
As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$
$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$
from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:
$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$
$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$
$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$