Hello, I know from UV/VIS spectroscopy that for many solutions a single frequency from the visible spectrum is absorbed. The color of such a solution is the color complementary to the one absorbed. If red is absorbed, then presumably it is no longer available to mix with its complementary green, so green is transmitted and the solution appears green. A colorless solution transmits all colors, so they are all subsequently still available to mix with each other to transmit the same white light that was incident on the solution. This leads me to believe that combinations of complementary colors make white. But this is inconsistent with what happens if I mix paint. Mixing two complementary colors makes brown, instead of white. If I mix all the primary and secondary colors together (as should occur with normal incident "white" light), I still get a mixture of browns. So my question is: Why is my solution colorless by mixing two complementary colors, but my paint brown? I suppose another way to ask the question might be if I hypothetically labeled the colors with numerical values. Colorless Solution: If red has a value of 2 and green a value of 3, then in my solution 2 + 3 = 5 and 5 is equal to white light. 5 - 3 = 2 and corresponds to absorption of green leaving red behind. Paint: If I mix red and green together, 2 + 3 = 5, where 5 is no longer equal to white light, rather reflects brown. I have no idea how to explain this or what is "complementary" to brown such that brown would be reflected. Thanks for any advice! I hope this question doesn't seem trivial or pointless, but the inconsistency between what happens in a solution and what is observed in paint makes it hard for me to wrap my head around the general concept.