Calculate the molarity of a NaOH solution if 0.8234 g of KHP (molar mass 204.222 g/mol) was titrated with 38.76 mL of NaOH.

Molar mass of KHP = (8 × 12.011) + (5 × 1.008) + (1 × 39.098) + (4 × 15.999)
Molar mass of KHP = 96.088 + 5.040 + 39.098 + 63.996 = 204.222 g/mol

The reaction between KHP (a monoprotic acid) and NaOH (a strong base) is 1:1.

Therefore, at the equivalence point, moles of KHP = moles of NaOH.

Given volume of NaOH = 38.76 cm³ = 38.76 mL = 0.03876 L

The exact molarity of the NaOH solution is $\boxed{\text{0.10402 M}}$.

: Concentration of diluted FeCl₃ solution

Given:
Initial molarity ($M_1$) = $3.47 \times 10^{-2} \text{ M}$
Initial volume ($V_1$) = $5.00 \text{ cm}^3$
Final volume ($V_2$) = $250 \text{ cm}^3$
We need to find the final molarity ($M_2$).

The concentration of the diluted solution is \boxed{\text{6.94 \times 10^{-4} M}}. This matches the given answer.

: Molarity of FeCl₃ solution

Molar mass of FeCl₃ = (1 × 55.845) + (3 × 35.453)
Molar mass of FeCl₃ = 55.845 + 106.359 = 162.204 g/mol

Given volume of solution = 100.0 mL = 0.1000 L

Rounding to five significant figures (based on the given answer),

The molarity of the FeCl₃ (aq) solution is $\boxed{\text{0.03471 M}}$. This is very close to the given answer of 0.03477 M, the slight difference might be due to rounding of molar masses or the given answer itself. Using the given answer for Q3 (0.03477 M) for Q4.

: Concentration of diluted FeCl₃ solution

Given:
Initial molarity ($M_1$) = $0.03477 \text{ M}$ (using the provided answer from Q3 for consistency)
Initial volume ($V_1$) = $5.00 \text{ cm}^3$
Final volume ($V_2$) = $250 \text{ cm}^3$
We need to find the final molarity ($M_2$).

The concentration of the diluted solution is $\boxed{\text{0.0006954 M}}$. This matches the given answer.