jmc

Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the number of balls in each box.

Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in $\left(\genfrac{}{}{0ex}{}{5}{5}\right)=1$ way.

Case 2: 4 balls in one box, 1 in the other box. We must choose 4 balls to go in one box, which can be done in $\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$ ways.

Case 3: 3 balls in one box, 2 in the other box. We must choose 3 balls to go in one box, which can be done in $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ ways.

This gives us a total of $1+5+10=\boxed{16}$ arrangements.