The chemist is using one solution that is 32% sodium and another solution is 12% sodium. How many liters of each type should the chemist use to produce 50 liters of the solution that is 20% sodium?

Let's denote the amount of the 32% solution as x and the amount of the 12% solution as y.

We know two things from the problem:

1. The total volume of the solution is 50 liters. This gives us the equation: x + y = 50.

2. The final solution is 20% sodium. This means that the amount of sodium from the 32% solution (0.32x) plus the amount of sodium from the 12% solution (0.12y) should make up 20% of the total 50 liters solution. This gives us the equation: 0.32x + 0.12y = 0.20 * 50.

Now we have a system of two equations, and we can solve it step by step:

First, let's multiply the first equation by 0.12 to make the coefficient in front of y the same in both equations:

0.12x + 0.12y = 0.12 * 50.

Now, subtract this equation from the second one:

0.32x + 0.12y - (0.12x + 0.12y) = 0.20 * 50 - 0.12 * 50,

0.20x = 0.08 * 50,

x = 0.08 * 50 / 0.20 = 20.

Substitute x = 20 into the first equation:

20 + y = 50,

y = 50 - 20 = 30.

So, the chemist should use 20 liters of the 32% solution and 30 liters of the 12% solution.