M1V1 = M2V2
M1 = 2 M
V1 = 2 L
V2 = 50 L
M2 = (M1V1)/V2
M2 = (2 M)(2 L)/(50 L)
M2 = 0.08 M or [H] = 0.08 mol/L.
water to make a final solution of 50 L What is the final molarity or [H] of the fish tank?
M1 = 2 M
V1 = 2 L
V2 = 50 L
M2 = (M1V1)/V2
M2 = (2 M)(2 L)/(50 L)
M2 = 0.08 M or [H] = 0.08 mol/L.
The initial volume and concentration of the acid are given as:
Volume (V1) = 2 L
Concentration (C1) = 2 M
The final volume of the solution after dilution is:
Volume (V2) = 50 L
Using the dilution formula, which states that the initial concentration times the initial volume is equal to the final concentration times the final volume, we can calculate the final molarity.
C1V1 = C2V2
(2 M) (2 L) = C2 (50 L)
Simplifying the equation:
4 M·L = 50 C2·L
Dividing both sides by 50 L:
C2 = (4 M·L) / 50 L
C2 = 0.08 M
Therefore, the final molarity or [H] of the fish tank is 0.08 M.
Dilution is the process of adding more solvent (in this case, water) to a solution to decrease the concentration of the solute (in this case, acid). The total moles of the solute before and after dilution should remain the same.
First, let's calculate the initial number of moles of acid in the 2 L of 2 M solution.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, 2 M means there are 2 moles of acid in 1 liter of solution. Since we have 2 liters of solution, the initial number of moles of acid is:
2 M * 2 L = 4 moles of acid
Now, let's calculate the final molarity or [H] of the fish tank after dilution.
Using the concept of dilution, we know that the number of moles of acid remains the same, but the volume of the solution increases from 2 L to 50 L. Therefore, the final molarity can be calculated as:
Final Molarity = Initial Moles of Acid / Final Volume of Solution
Final Molarity = 4 moles of acid / 50 L = 0.08 M
Therefore, the final molarity or [H] of the fish tank is 0.08 M.