Question 3.EP.11: A chemist requires 1.5 M hydrochloric acid, HCl, for a serie...
A chemist requires 1.5 M hydrochloric acid, HCl, for a series of reactions. The only solution available is 6.0 M HCl. What volume of 6.0 M HCl must be diluted to obtain 5.0 L of 1.5 M HCl?
Strategy
First we need to recognize that this is the dilution of the concentrated solution to prepare the desired one. The underlying concept we must use is that the number of moles of HCl will be the same before and after dilution, and this means we can use Equation 3.3.
M_{i}\times V_{i} = M_{f}\times V_{f} (3.3)
We know the desired final molarity and final volume, as well as the initial molarity. So we can solve for the needed initial volume.
Initial concentration of HCl: M_{i} = 6.0 M
Final concentration of HCl: M_{f} = 1.5 M
Final volume of solution: V_{f} = 5.0 L
The unknown is the initial volume, V_{i}. Rearranging Equation 3.3,
V_{i} = \frac{M_{f}\times V_{f}}{M_{i}}
Inserting the known quantities on the right-hand side,
V_{i} = \frac{1.5 \ M\times 5.0 \ L}{6.0 \ M}= 1.3 \ L
To obtain the desired quantity of diluted HCl, the chemist should begin with 1.3 L of the concentrated solution and add enough water to bring the volume up to 5.0 L.
Analyze Your Answer
Once we are certain that we are dealing with a dilution problem, one way to make sure the answer makes sense is to think about which solution is concentrated and which is dilute. You will always need smaller volumes of the concentrated solution. Did we get it right this time? It seems likely that we did: we were asked to calculate how much of the concentrated solution we would need, so the volume we calculate should be the smaller of the two volumes involved in the problem.