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

To find the volume of water added to the NaOH solution, we need to subtract the volume of the resulting solution from the initial volume of NaOH solution.

Given:
Initial volume of NaOH solution (V1) = 310 mL

We know that volumes are additive, so the volume of the resulting solution (V2) is the sum of the initial NaOH solution and the volume of water added.

Let's represent the volume of water added as x mL.

V2 = V1 + x

Since we don't know the volume of the resulting solution, we'll use the fact that 75 mL of the resulting solution was titrated with 58.2 mL of 6 M HCl. This indicates that the volume of the resulting solution is 75 mL.

75 mL = 310 mL + x

Now we can solve for x by subtracting 310 mL from both sides:

75 mL - 310 mL = x

x = -235 mL

The negative value of x indicates an error since we can't have a negative volume of water. Therefore, there might be a mistake in the given information or in the calculation. Please recheck the question or provide additional information if possible.

To find the volume of water added to the 310 mL of NaOH solution, we need to calculate the volume of the resulting solution after the water is added.

We know that the final volume of solution = initial volume of NaOH solution + volume of water.

Let's assume the volume of water added is 'x' mL.

Therefore, the final volume of solution = 310 mL + x mL.

Next, we are given that a 75 mL sample of the resulting solution is titrated to neutrality with 58.2 mL of 6 M HCl.

This means that the moles of HCl reacting with the NaOH in the sample are equal, as they react in a 1:1 ratio in a neutralization reaction.

Since the HCl solution has a concentration of 6 M and a volume of 58.2 mL, the number of moles of HCl present in the titration can be calculated as follows:

moles of HCl = concentration of HCl × volume of HCl solution in liters
= 6 M × 0.0582 L (since 58.2 mL = 0.0582 L)
= 0.3492 moles

Since the reaction between NaOH and HCl is 1:1, the number of moles of NaOH present in the 75 mL sample also = 0.3492 moles.

Now, we can calculate the concentration of the diluted NaOH solution:

concentration of diluted NaOH solution = moles of NaOH / final volume of solution in liters

Since the moles of NaOH = 0.3492 moles and the final volume of solution = 310 mL + x mL (which needs to be converted to liters), we have:

concentration of diluted NaOH solution = 0.3492 moles / (0.31 L + x L)

To express the answer in units of M (moles per liter), we can simplify the calculation by converting the initial volume from mL to L:

concentration of diluted NaOH solution = 0.3492 moles / ((310 mL + x mL) / 1000)

Therefore, the concentration of the diluted NaOH solution is 0.3492 M / ((310 mL + x mL) / 1000).

To find the volume of water added, we need to solve the equation when the resulting solution volume is substituted by the titration volume:

0.3492 M / ((310 mL + x mL) / 1000) = 0.3492 M / ((75 mL + 58.2 mL) / 1000)

Now, we can solve for 'x' by cross multiplying:

(310 mL + x mL) / 1000 = (75 mL + 58.2 mL) / 1000

310 mL + x mL = 133.2 mL

x mL = 133.2 mL - 310 mL = -176.8 mL

The negative value of 'x' indicates that there was an excess of 176.8 mL of water in the solution, which means 176.8 mL of water was added to the 310 mL of NaOH solution.

Therefore, the volume of water added to the 310 mL of NaOH solution is 176.8 mL.