In the following reaction

$\underset{\left(A\right)}{{\left[Cu{\left(H_{2}O\right)}_3\left(OH\right)\right]}^{+}}+\underset{\left(B\right)}{{\left[Al{\left(H_{2}O\right)}_6\right]}^{3+}}\stackrel{}{\to }\underset{\left(C\right)}{{\left[Cu{\left(H_{2}O\right)}_4\right]}^{2+}}+\underset{\left(D\right)}{{\left[Al{\left(H_{2}O\right)}_5\left(OH\right)\right]}^{2+}}$

$\underset{\left(\mathrm{A}\right)}{{\left[\mathrm{Cu}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3\left(\mathrm{OH}\right)\right]}^{+}}+\underset{\left(\mathrm{B}\right)}{{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}}\stackrel{ }{\to }\underset{\left(\mathrm{C}\right)}{{\left[\mathrm{Cu}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\right]}^{2+}}+\underset{\left(\mathrm{D}\right)}{{\left[\mathrm{Al}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\left(\mathrm{OH}\right)\right]}^{2+}}$

A accepts H⁺ from B, therefore A is base and B is and acid.

Conjugate acid is the substance which is formed when base accepts H⁺ and conjugate base is the substance which is formed by donation of H⁺ by acid.

So, (C) is the conjugate acid of (A) and (D) is the conjugate base of (B).