Balance the reaction in basic medium. (6pts) K2Cr2O7 + Na2SO3 + H2O → Cr(OH)3 + Na2SO4 + KOH

First, let's write down the unbalanced chemical equation:

K_2Cr_2O_7 + Na_2SO_3 + H_2O \rightarrow Cr(OH)_3 + Na_2SO_4 + KOH

Next, we need to balance the equation by making sure that the number of atoms of each element is the same on both sides of the equation.

1. Balance the elements other than hydrogen and oxygen:
K_2Cr_2O_7 + 3Na_2SO_3 + H_2O \rightarrow 2Cr(OH)_3 + 3Na_2SO_4 + 2KOH

2. Balance the oxygen atoms by adding water molecules:
K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2O \rightarrow 2Cr(OH)_3 + 3Na_2SO_4 + 2KOH

3. Balance the hydrogen atoms by adding hydroxide ions (OH⁻):
K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2O \rightarrow 2Cr(OH)_3 + 3Na_2SO_4 + 2KOH

Therefore, the balanced chemical equation in basic medium is:

\boxed{K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2O \rightarrow 2Cr(OH)_3 + 3Na_2SO_4 + 2KOH}