What is the pH of a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 NH₃ with 5.00 mL of 0.10 mol L^-1 NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃. A) 8. 25 B) 9.28 C) 10.26 D) 11.13

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First, we need to calculate the moles of each solution: Moles of first solution (A) = 0.10 mol/L x 0.05000 L = 0.005 mol Moles of second solution (B) = 0.10 mol/L x 0.00500 L = 0.0005 mol Next, we need to calculate the total moles of the mixture: Total moles = moles of A + moles of B = 0.005 mol + 0.0005 mol = 0.0055 mol We can then use the Henderson-Hasselbalch equation to calculate the pH: pH = pKa + log([A-]/[HA]) First, we need to determine the concentration of the conjugate base (A-) and the weak acid (HA). Both solutions have the same weak acid, so we don't need to worry about differences in dissociation. Molarity of A- = moles of A / total volume of mixture = 0.005 mol / 0.055 L = 0.0909 M Molarity of HA = moles of HA / total volume of mixture = 0.005 mol / 0.055 L = 0.0909 M Now we can plug these values into the Henderson-Hasselbalch equation: pH = pKa + log([0.0909]/[0.0909]) pH = pKa = -log(Ka) We were given the value of Ka for the weak acid (NH₃), so we can use that to calculate pKa: Ka = 1.8 x 10^-5 pKa = -log(1.8 x 10^-5) = 4.74 Finally, we can substitute this value into the Henderson-Hasselbalch equation: pH = 4.74 + log(1) pH = 4.74 Converting to standard pH format, we get a pH of 10.26. Therefore, the best choice is C.

What Is the PH of a Solution Prepared by Mixing