Solving a System of Equations from a Mixture Problem
A chemistry student is given a 100.0 mL mixture of two solutions: one containing 0.500 M HCl and the other containing 0.300 M HCl. The student is told that the total amount of HCl in the mixture is 0.0420 moles. The student needs to determine the volume of each solution used to make the mixture. Let x be the volume (in liters) of the 0.500 M HCl solution and y be the volume (in liters) of the 0.300 M HCl solution. The student sets up the following system of equations: x + y = 0.1000 0.500x + 0.300y = 0.0420 After solving, the student obtains x = 0.0600 L and y = 0.0400 L. The student then checks the answer by calculating the total moles and volume.
1. Solve the system of equations algebraically. Show your steps and verify the solution.
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From the first equation: y = 0.1000 - x. Substitute into the second: 0.500x + 0.300(0.1000 - x) = 0.0420. 0.500x + 0.0300 - 0.300x = 0.0420. 0.200x = 0.0120. x = 0.0600 L. Then y = 0.1000 - 0.0600 = 0.0400 L. Check: total volume = 0.0600 + 0.0400 = 0.1000 L. Total moles = 0.500(0.0600) + 0.300(0.0400) = 0.0300 + 0.0120 = 0.0420 mol. The solution is correct.
2. If the student had mistakenly used 0.400 M instead of 0.300 M for the second solution, what would be the new volumes? Would the total moles still be 0.0420?
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New system: x + y = 0.1000; 0.500x + 0.400y = 0.0420. Substitute y = 0.1000 - x: 0.500x + 0.400(0.1000 - x) = 0.0420. 0.500x + 0.0400 - 0.400x = 0.0420. 0.100x = 0.0020 → x = 0.0200 L, y = 0.0800 L. Total moles = 0.500(0.0200) + 0.400(0.0800) = 0.0100 + 0.0320 = 0.0420 mol. Yes, the total moles remain the same because the equation is still satisfied, but the volumes change.