Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make 250.0 \(\mathrm{mL}\) of solution having a pH of 10.00\(\left(K_{\mathrm{b}}\right.\) \(=1.1 \times 10^{-8} )\)

The relationship between pH and pOH is: pH + pOH = 14 We are given pH = 10, so: pOH = 14 - 10 = 4

We can determine the concentration of OH⁻ ions from the pOH using the following formula: \[ OH^⁻ = 10^{-pOH} \] \[ OH^⁻ = 10^{-4} = 1 \times 10^{-4} \, M\]

Let's represent the dissociation of HONH₂ as: HONH₂ (aq) + H₂O (l) ⇌ HONH⁻ (aq) + OH⁻ (aq) The Kb expression can be written as: \[ K_b = \frac{[HONH⁻][OH-]}{[HONH_2]} \] We know the Kb value and the concentration of OH⁻. Assuming x mol/L is the concentration of HONH₂ in equilibrium, we have: \[ 1.1 \times 10^{-8} = \frac{x(1 \times 10^{-4})}{x - 1 \times 10^{-4}} \] Now we can solve for x, which represents the equilibrium concentration of HONH₂. The concentration of HONH₂ is very small compared to the initial concentration; hence x - 1 × 10^(-4) ≈ x \[ 1.1 \times 10^{-8} = \frac{x \times 1 \times 10^{-4}}{x} \] \[ x = 1.1 \times 10^{-4} \, M\]

Now that we have the concentration of HONH₂, we can calculate the mass required. We know that the volume of the solution is 250.0 mL. First, let's convert the volume into liters: Volume = 250.0 mL × (1 L / 1000 mL) = 0.250 L Next, we can calculate the moles of HONH₂ using the concentration: Moles of HONH₂ = concentration × volume Moles of HONH₂ = (1.1 × 10⁻⁴ M) × (0.250 L) = 2.75 × 10⁻⁵ mol Finally, we can determine the mass by multiplying the moles by the molecular weight of HONH₂ (Molar mass of HONH₂ = 1(H) + 15(N) + 16(O) + 1(H) + 1(N) + 2(H) = 33 g/mol): Mass of HONH₂ = moles × molecular weight Mass of HONH₂ = (2.75 × 10⁻⁵ mol) × (33 g/mol) ≈ 0.0009075 g Therefore, the mass of HONH₂ required to make a 250.0 mL solution with a pH of 10.00 is approximately 0.0009 g (rounded to four decimal places).