Let x liters of a 2 % HCL is mixed with (3-x) liters of 11 % HCL solution to obtain 3 liters of 8% HCL solution. So the equation will be
[tex]0.02x + 0.11(3-x)= 3*0.08[/tex]
Distributing 0.11 on 3-x, we will get
[tex]0.02x +0.33 - 0.11x = 0.24[/tex]
Combining like terms
[tex]-0.09x = -0.09[/tex]
Dividing both sides by -0.09
[tex]x=1 liter[/tex]
So 1 liter of 2 % HCL solution is mixed with 2 liters of 11% HCl solution to get 3 liters of 8% HCL solution .