An unknown amount of water is mixed with 310 mL of a 6 M solution of NaOH solution. A 75 mL sample of the resulting solution is titrated to neutrality with 58.2 mL of 6 M HCl. Calculate the concentration of the diluted NaOH solution. Answer in units of M
What volume of water was added to the 310 mL of NaOH solution? Assume volumes are additive. Answer in units of mL
M of the 75 mL =
58.2 mL x 6M/75 mL = 4.656M
So the initial solution must have been
6M NaOH x 310 mL = 4.656M x ? mL
Solve for ? mL, then subract 310 from it to see how much water was added.
To find the volume of water added to the NaOH solution, we can subtract the known volume of the 6 M NaOH solution from the total volume of the resulting solution.
Total volume of the resulting solution = volume of 6 M NaOH solution + volume of water
Given:
Volume of the resulting solution = unknown
Volume of 6 M NaOH solution = 310 mL
Let's assume the volume of water added is 'x' mL.
Therefore,
Total volume of the resulting solution = 310 mL + x mL
Since the problem states that the volume of water added is unknown, we can use a variable 'x' to represent it. Now we can plug this into the equation:
Total volume of the resulting solution = 310 mL + x mL
To solve for 'x', we need additional information.