A 0.401 g sample of \( \mathrm{KHP}\left(\mathrm{HKC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}, \mathrm{MM}=204.23 \mathrm{~g} / \mathrm{mol}\right) \) required 31.00 mL of NaOH solution to reach a phenolphthalein endpoint. Determine the molarity of the NaOH : (A) \( 0.0633 \mathrm{~mol} / \mathrm{L} \) (B) \( 0.63 \mathrm{~mol} / \mathrm{L} \) (C) \( 2.64 \mathrm{~mol} / \mathrm{L} \) (D) \( 15.8 \mathrm{~mol} / \mathrm{L} \) (E) \( 16 \mathrm{~mol} / \mathrm{L} \)

To determine the molarity of the NaOH solution, we first need to calculate the number of moles of KHP in the 0.401 g sample. 1. Calculate moles of KHP: \[ \text{Moles of KHP} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.401 \text{ g}}{204.23 \text{ g/mol}} \approx 0.00196 \text{ mol} \] 2. KHP reacts with NaOH in a 1:1 ratio. Therefore, the moles of NaOH will also be 0.00196 mol. 3. Calculate molarity of NaOH: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} \] The volume of NaOH is 31.00 mL, which is 0.03100 L. \[ \text{Molarity of NaOH} = \frac{0.00196 \text{ mol}}{0.03100 \text{ L}} \approx 0.06323 \text{ mol/L} \] So, the molarity of the NaOH solution is approximately \( 0.0633 \, \text{mol/L} \), which corresponds to option (A). Now for some additional fun facts: Did you know that KHP (Potassium Hydrogen Phthalate) is commonly used as a primary standard in acid-base titrations because it is stable, non-hygroscopic, and easily soluble in water? This makes it perfect for accurately determining concentrations of solutions, such as NaOH, in a laboratory! Additionally, this reaction between KHP and NaOH is a classic example of a neutralization reaction, where an acid (KHP) and a base (NaOH) combine to form water and a salt. Neutralization reactions are all around us—just think about making a delicious lemonade by balancing the sweetness of sugar (acidic) and the tangy sourness of lemons (basic)!