Consider the titration of 300.0 mL of 0.577 M NH₃ (K_b = 1.8 $\times$ 10^-⁵)with 0.500 M HNO₃.After 150.0 mL of 0.500 M HNO₃ have been added,the pH of the solution is:

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This is a titration of a weak acid (NH₃) with a strong base (HNO₃). When the titrant (0.500 M HNO₃) is added to the weak acid solution, it will react with the acid to form its conjugate base (_b-). Initially, the pH of the solution will be acidic due to the presence of the weak acid. As the titrant is added, the pH will increase until it reaches the equivalence point, which is when all of the weak acid has been neutralized and the solution contains only the conjugate base. At the equivalence point, the pH will be equal to the pKa of the weak acid. The pKa of NH₃ is given as 10.10, so the pH at the equivalence point is expected to be around 10.10. Since we have added less than the equivalence point volume of the titrant (150.0 mL of 0.500 M HNO₃ to 300.0 mL of 0.577 M NH₃), the pH will be less than the pKa of the weak acid. Using the Henderson-Hasselbalch equation: pH = pKa + log([_b-]/[NH₃]) At the halfway point of the titration, where the volume of titrant added is equal to half of the equivalence point volume, we can assume that half of the weak acid has been neutralized and converted to its conjugate base. Therefore, we can use the following approximate values: [_b-] = [NH₃] pH = pKa + log(1) pH = pKa pH = 10.10 Since we have added less than the halfway point volume of titrant, the pH will be less than 10.10. Looking at the answer choices, the only option that is less than 10.10 is 9.37, which is choice D. Therefore, the answer is D.

Consider the Titration of 300 ×\times× 10-5)with 0

Consider the Titration of 300 $\times$ 10^-⁵)with 0