50.0mL of 0.200M HCl diluted to 100.0 mL and titrated with 0.200M NaOH. what is total volume of solution(L) if volume of NaOH added(mL) are 0.00, 5.00, and 10.0
Isn't this just 100 mL + 0, 100 + 5 and 100 + 10? or have I missed something.
To find the total volume of the solution in liters, we need to consider the initial volume of HCl and the volume of NaOH added in each case.
Given:
Initial volume of HCl (V1) = 50.0 mL
Final volume of the solution (V2) = 100.0 mL
Volume of NaOH added in each case = 0.00 mL, 5.00 mL, 10.0 mL
To calculate the total volume of the solution in liters, we need to convert all volumes to liters and add them together.
Converting the volumes to liters:
V1 = 50.0 mL = 50.0 mL * (1 L / 1000 mL) = 0.050 L
V2 = 100.0 mL = 100.0 mL * (1 L / 1000 mL) = 0.100 L
Volume of NaOH added in each case:
0.00 mL = 0.00 mL * (1 L / 1000 mL) = 0.000 L
5.00 mL = 5.00 mL * (1 L / 1000 mL) = 0.005 L
10.0 mL = 10.0 mL * (1 L / 1000 mL) = 0.010 L
Now, let's calculate the total volume of the solution for each case:
Case 1: Volume of NaOH added = 0.00 mL
Total volume = V2
Total volume = 0.100 L
Case 2: Volume of NaOH added = 5.00 mL
Total volume = V1 + Volume of NaOH added
Total volume = 0.050 L + 0.005 L
Total volume = 0.055 L
Case 3: Volume of NaOH added = 10.0 mL
Total volume = V1 + Volume of NaOH added
Total volume = 0.050 L + 0.010 L
Total volume = 0.060 L
Therefore, the total volume of the solution in liters for each case is:
Case 1: 0.100 L
Case 2: 0.055 L
Case 3: 0.060 L