A student must make 500 mL of a 1.5 M HCl solution. The student has plenty of water and a stock solution of 6.0 M HCl to use. What volume of the 6.0 M HCl solution must be added to what volume of water to make the desired 500 mL of 1.5 M solution?
I don't really know how to start this. I've tried a couple of ways. I got that you need .75 moles of HCl and therefore you need .125 L of the 6.0 M solution, but I don't really think that that's correct and if it is I don't know what to do to find what volume of water you would need.
Here is another way: You want to dilute it 4 times, so use one part 6M acid, three parts water.
dividing 500mL by four parts is indeed one part 125mL of the acid, and then three parts water.
Thanks! That makes a lot more sense.
Molarity of the first and volume of the first = molarity of the 2nd
To solve this problem, we need to use the concept of molarity and the equation for dilution. Here's how you can obtain the answer.
First, let's determine the number of moles of HCl we need for the desired solution. We know that the desired concentration is 1.5 M, and the desired volume is 500 mL (or 0.5 L).
Moles of HCl = concentration × volume
Moles of HCl = 1.5 M × 0.5 L
Moles of HCl = 0.75 moles
As you correctly calculated, we need 0.75 moles of HCl for the solution.
Now, let's determine the volume of the 6.0 M HCl solution required. To do that, we can rearrange the equation for molarity:
Moles of solute = Molarity × Volume of solution
In this case, the solute is the 6.0 M HCl solution, and we need to solve for the volume.
Volume of 6.0 M HCl = Moles of HCl ÷ Molarity of HCl
Volume of 6.0 M HCl = 0.75 moles ÷ 6.0 M
Volume of 6.0 M HCl = 0.125 L or 125 mL
So, you were correct that 0.125 L or 125 mL of the 6.0 M HCl solution is needed.
Finally, to find the volume of water required, we can subtract the volume of the 6.0 M HCl solution from the total desired volume:
Volume of water = Total volume - Volume of 6.0 M HCl
Volume of water = 500 mL - 125 mL
Volume of water = 375 mL
Therefore, to make 500 mL of a 1.5 M HCl solution, you need to add 125 mL of the 6.0 M HCl solution to 375 mL of water.