Calculate the pH of a solution made by mixing 100.0 mL of 0.635 M NH₃ with 100.0 mL of 0.100 M HCl.(K_b for NH₃ = 1.8 $\times$ 10^-5) A) 9.98 B) 4.02 C) 8.53 D) 9.26 E) none of these

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The correct answer is A because the pH of the solution is 9.98. This can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of the weak acid (NH3), [A-] is the concentration of the conjugate base (NH4+), and [HA] is the concentration of the weak acid (NH3). In this case, pKa = -log(Kb) = -log(1.8 x 10^-5) = 4.74, [A-] = [NH4+] = 0.100 M, and [HA] = [NH3] = 0.635 M - 0.100 M = 0.535 M. Plugging these values into the Henderson-Hasselbalch equation gives pH = 4.74 + log(0.100/0.535) = 9.98.

Calculate the PH of a Solution Made by Mixing 100 ×\times×

Calculate the PH of a Solution Made by Mixing 100 $\times$