Color blending is just a linear interpolation per channel, right? So the math is pretty simple. If you have RGBA1 over RGB2, the effective visual result RGB3 will be:
r3 = r2 + (r1-r2)*a1
g3 = g2 + (g1-g2)*a1
b3 = b2 + (b1-b2)*a1
…where the alpha channel is from 0.0 to 1.0.
Sanity check: if the alpha is 0, is RGB3 the same as RGB2? Yes. If the alpha is 1, is RGB3 the same as RGB1? Yes.
If you locked down only the background color and final color, there are a large number of RGBA colors (infinite, in floating-point space) that could satisfy the requirements. So you have to pick either the color of the bar or the opacity level you want, and find out the value of the other.
Picking the Color Based on Alpha
If you know RGB3 (the final desired color), RGB2 (the background color), and A1 (how much opacity you want), and you are just looking for RGB1, then we can re-arrange the equations thusly:
r1 = (r3 - r2 + r2*a1)/a1
g1 = (g3 - g2 + g2*a1)/a1
b1 = (b3 - b2 + b2*a1)/a1
There are some color combinations which are theoretically possible, but impossible given the standard RGBA range. For example, if the background is pure black, the desired perceived color is pure white, and the desired alpha is 1%, then you would need:
r1 = g1 = b1 = 255/0.01 = 25500
…a super-bright white 100× brighter than any available.
Picking the Alpha Based on Colors
If you know RGB3 (the final desired color), RGB2 (the background color), and RGB1 (the color you have that you want to vary the opacity of), and you are just looking for A1, then we can re-arrange the equations thusly:
a1 = (r3-r2) / (r1-r2)
a1 = (g3-g2) / (g1-g2)
a1 = (b3-b2) / (b1-b2)
If these give different values, then you can't make it match exactly, but you can average the alphas to get as close as possible. For example, there's no opacity in the world that will let you put green over red to get blue.