5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$

be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$
5. -2 .

From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$, and for any $x \in\left(\frac{1}{2}, 1\right)$, $f^{\prime}(x) > 0$, so
$$
f(x)_{\text {min }}=f\left(\frac{1}{2}\right)=-1 .
$$

Let $x_{1}+x_{2}=x \in(0,1)$. Then
$$
x_{3}+x_{4}=1-x \in(0,1) \text {. }
$$

From $\frac{x_{1}}{x}+\frac{x_{2}}{x}=1$, we get
$$
\frac{x_{1}}{x} \log _{2} \frac{x_{1}}{x}+\frac{x_{2}}{x} \log _{2} \frac{x_{2}}{x}=f\left(\frac{x_{1}}{x}\right) \geqslant-1,
$$

which means $x_{1} \log _{2} x_{1}+x_{2} \log _{2} x_{2} \geqslant-x+x \log _{2} x$.
Similarly, from the above, we get
$$
\begin{array}{l}
x_{3} \log _{2} x_{3}+x_{4} \log _{2} x_{4} \\
\geqslant-(1-x)+(1-x) \log _{2}(1-x) .
\end{array}
$$

Adding the two inequalities, we get
$$
\sum_{i=1}^{4} x_{i} \log _{2} x_{i} \geqslant-1+f(x) \geqslant-2 \text {. }
$$

When $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{4}$, the equality holds.
Thus, $s_{\min }=-2$.