If 2.29 mL of a stock solution of acetic acid with a concentration of 0.723 M is mixed with 15.75 mL of water, what is the concentration of the diluted solution?
volume to start = 2.29
volume added = 15.75
total volume = ?
0.723M x (2.29/total volume) = ?
Note: This assumes volumes are additive; technically they are not but in dilute solutions like this the are for all practical purposes (still not technically)
To find the concentration of the diluted solution, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (0.723 M)
V1 = volume of the stock solution (2.29 mL)
C2 = concentration of the diluted solution (unknown)
V2 = volume of the diluted solution (15.75 mL)
First, convert the volumes to liters:
V1 = 2.29 mL / 1000 mL/L = 0.00229 L
V2 = 15.75 mL / 1000 mL/L = 0.01575 L
Now, substitute the values into the formula:
(0.723 M)(0.00229 L) = C2(0.01575 L)
Solving for C2:
C2 = (0.723 M)(0.00229 L) / 0.01575 L
Calculating the value:
C2 ≈ 0.105 M
Therefore, the concentration of the diluted solution is approximately 0.105 M.