A student dissolves 4.00 g of an unknown monoprotic acid in 1.00 L of water. They titrate this solution with 0.600 M NaOH, and the equivalence point is reached when 38.7 mL of NaOH is added. Calculate

The volume of NaOH is $38.7 \text{ mL} = 0.0387 \text{ L}$.
The molarity of NaOH is $0.600 \text{ M}$.

$$\text{Moles of NaOH} = \text{Molarity} \times \text{Volume}$$ $$\text{Moles of NaOH} = 0.600 \frac{\text{mol}}{\text{L}} \times 0.0387 \text{ L} = 0.02322 \text{ mol}$$

A monoprotic acid (HA) reacts with NaOH in a 1:1 mole ratio:

$$\text{HA(aq)} + \text{NaOH(aq)} \longrightarrow \text{NaA(aq)} + \text{H}_2\text{O(l)}$$

Since $0.02322 \text{ mol}$ of NaOH was needed to neutralize "the acid", we assume this refers to the total moles of acid present in the solution.

$$\text{Moles of HA} = \text{Moles of NaOH} = 0.02322 \text{ mol}$$

The $0.02322 \text{ mol}$ of acid was dissolved in $1.00 \text{ L}$ of water.

$$\text{Molarity of acid} = \frac{\text{Moles of HA}}{\text{Volume of solution}}$$ $$\text{Molarity of acid} = \frac{0.02322 \text{ mol}}{1.00 \text{ L}} = 0.02322 \text{ M}$$

Rounding to three significant figures:

$$\text{Molarity of acid} = \boxed{\text{0.0232 M}}$$

The mass of the unknown acid is $4.00 \text{ g}$.

$$\text{Molar mass of acid} = \frac{\text{Mass of acid}}{\text{Moles of acid}}$$ $$\text{Molar mass of acid} = \frac{4.00 \text{ g}}{0.02322 \text{ mol}} \approx 172.26 \frac{\text{g}}{\text{mol}}$$

Rounding to three significant figures:

$$\text{Molar mass of acid} = \boxed{\text{172 g/mol}}$$

Selenium (Se) is element 34. Its neutral electron configuration is $[\text{Ar}] 3d^{10} 4s^2 4p^4$.
To form Se²⁻, two electrons are gained, filling the $4p$ subshell.

$$\text{Se}^{2-} : \boxed{[\text{Ar}] 3d^{10} 4s^2 4p^6}$$

Iron (Fe) is element 26. Its neutral electron configuration is $[\text{Ar}] 3d^6 4s^2$.
To form Fe²⁺, two electrons are lost. For transition metals, electrons are removed from the highest principal energy level first, which is the $4s$ subshell.

$$\text{Fe}^{2+} : \boxed{[\text{Ar}] 3d^6}$$