In a acid-base titration, 22.81 mL of an NaOH solution were required to neutralize 26.18 mL of a 0.1121 M HCl solution. What is the molarity of the NaOH solution?
Write the equation to prove to yourself that it is 1 mole acid reacting with 1 mole base. That so, then
M1V1 = M2V2.
You have one unknown. Solve for that.
To find the molarity (M) of the NaOH solution, we can use the equation:
M1V1 = M2V2
Where:
M1 is the molarity of the HCl solution
V1 is the volume of the HCl solution in liters
M2 is the molarity of the NaOH solution
V2 is the volume of the NaOH solution in liters
Let's plug in the given values into the equation:
M1 = 0.1121 M
V1 = 26.18 mL = 26.18 / 1000 L = 0.02618 L
V2 = 22.81 mL = 22.81 / 1000 L = 0.02281 L
Now, let's solve for M2:
M1V1 = M2V2
0.1121 M * 0.02618 L = M2 * 0.02281 L
0.00293 = M2 * 0.02281 L
M2 = 0.00293 / 0.02281 L
M2 = 0.128 M
Therefore, the molarity of the NaOH solution is 0.128 M.