An unknown amount of water is mixed with

350 mL of a 6 M solution of NaOH solution.
A 75 mL sample of the resulting solution is
titrated to neutrality with 51.2 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
350 mL of NaOH solution? Assume volumes
are additive. Answer in units of mL

This is the corrected copy. An equal sign I typed as a - sign.

millimoles HCl = 6M x 51.2 = 3072.
mmoles NaOH solution (75 mL) = 3072.
M NaOH soln = mmoles/mL = 3072/75 = 4.096M That's the molarity of the diluted solution.

The diluted solution started out as 6M and it was diluted and is now 4.096M. Let x = mL water added to the 6M NaOH.
6M x [350/(350+x) = 4.096
I calculated x to be approximately 163 mL so the final volume would be about 350 + 163 or about 513 mL. You need to clean up the numbers and watch the number of significant figures but this is the way you do it.

To solve this problem, we need to use the concept of molarity and the balanced chemical equation between NaOH and HCl.

Let's break down the steps to find the concentration of the diluted NaOH solution:

Step 1: Calculate the number of moles of HCl used in the titration.
We know the concentration and volume of the HCl used:
Concentration of HCl = 6 M
Volume of HCl = 51.2 mL = 0.0512 L

To calculate the number of moles, we use the formula:
Moles of solute = Concentration × Volume

Moles of HCl = 6 M × 0.0512 L

Step 2: Determine the balanced chemical equation between NaOH and HCl.
The balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl → NaCl + H2O

From the balanced equation, we know that 1 mole of NaOH reacts with 1 mole of HCl.

Step 3: Calculate the number of moles of NaOH that remained after the titration.
Since 1 mole of NaOH reacts with 1 mole of HCl, the number of moles of HCl used in the titration is also equal to the number of moles of NaOH. Therefore, the number of moles of NaOH that remained is equal to the moles of HCl used.

Number of moles of NaOH = Moles of HCl used

Step 4: Calculate the concentration of the diluted NaOH solution.
Now, we can calculate the concentration of the diluted NaOH solution.

We have the initial volume of the NaOH solution (350 mL) and the volume of water added to it. Let’s call the volume of water added "x" mL.

The final volume of the solution after adding water is (350 + x) mL.

Since the number of moles of NaOH that remained after the titration is the same as the number of moles of HCl used, we can set up the following equation:

Initial moles of NaOH = Final moles of NaOH

(6 M) × (0.075 L) = Concentration of diluted NaOH solution × (350 + x) mL

Solving for the concentration of the diluted NaOH solution:

Concentration of diluted NaOH solution = (6 M) × (0.075 L) / (350 + x) mL

Now, given that the concentration of the diluted NaOH solution is in units of M, the answer should be written as the concentration in moles per liter (M). To convert the volume of water added to mL (milliliters), we can directly use the given volume in mL.

Therefore, the concentration of the diluted NaOH solution is:

Concentration of diluted NaOH solution = (6 M) × (0.075 L) / (350 + x) mL

To find the volume of water added to the 350 mL of NaOH solution, you would need to solve for x in the above equation.