91影视

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鈦� ions from the pOH using the following formula: \[ OH^鈦� = 10^{-pOH} \] \[ OH^鈦� = 10^{-4} = 1 \times 10^{-4} \, M\]

Let's represent the dissociation of HONH鈧� as: HONH鈧� (aq) + H鈧侽 (l) 鈬� HONH鈦� (aq) + OH鈦� (aq) The Kb expression can be written as: \[ K_b = \frac{[HONH鈦籡[OH-]}{[HONH_2]} \] We know the Kb value and the concentration of OH鈦�. Assuming x mol/L is the concentration of HONH鈧� 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 HONH鈧�. The concentration of HONH鈧� is very small compared to the initial concentration; hence x - 1 脳 10^(-4) 鈮� 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 HONH鈧�, 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 脳 (1 L / 1000 mL) = 0.250 L Next, we can calculate the moles of HONH鈧� using the concentration: Moles of HONH鈧� = concentration 脳 volume Moles of HONH鈧� = (1.1 脳 10鈦烩伌 M) 脳 (0.250 L) = 2.75 脳 10鈦烩伒 mol Finally, we can determine the mass by multiplying the moles by the molecular weight of HONH鈧� (Molar mass of HONH鈧� = 1(H) + 15(N) + 16(O) + 1(H) + 1(N) + 2(H) = 33 g/mol): Mass of HONH鈧� = moles 脳 molecular weight Mass of HONH鈧� = (2.75 脳 10鈦烩伒 mol) 脳 (33 g/mol) 鈮� 0.0009075 g Therefore, the mass of HONH鈧� required to make a 250.0 mL solution with a pH of 10.00 is approximately 0.0009 g (rounded to four decimal places).