A 14.46 mL solution of acetic acid is titrated with a 0.250 M solution of NaOH . If 17.94 mL of the NaOH solution is needed to reach the end point, what is the molar concentration of the acetic acid solution? Report your answer with 3 significant figures, and do not include units in your answer.

To find the molar concentration of the acetic acid solution, we can start by calculating the moles of NaOH used in the titration. The moles of NaOH can be calculated using the formula: \[ \text{Moles of NaOH} = \text{Concentration of NaOH} \times \text{Volume of NaOH (in L)} \] First, convert the volume of NaOH from mL to L: \[ 17.94 \, \text{mL} = 0.01794 \, \text{L} \] Now, plug into the formula: \[ \text{Moles of NaOH} = 0.250 \, \text{mol/L} \times 0.01794 \, \text{L} = 0.004485 \, \text{mol} \] Since acetic acid (CH₃COOH) reacts with NaOH in a 1:1 molar ratio, the moles of acetic acid will also be 0.004485 mol. Next, we determine the molarity of the acetic acid solution using the formula: \[ \text{Molarity} = \frac{\text{Moles of acetic acid}}{\text{Volume of acetic acid (in L)}} \] Convert the volume of acetic acid from mL to L: \[ 14.46 \, \text{mL} = 0.01446 \, \text{L} \] Now we can calculate its concentration: \[ \text{Molarity} = \frac{0.004485 \, \text{mol}}{0.01446 \, \text{L}} \approx 0.310 \] So, the molar concentration of the acetic acid solution is 0.310.