Consider the titration of 300.0 mL of 0.569 M NH₃ (K_b = 1.8 $\times$ 10^-5)with 0.500 M HNO₃.After 150.0 mL HNO₃ has been added,what is the pH of the solution? A) 4.64 B) 9.36 C) 6.36 D) 11.36 E) 7.00

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Quizplus · Accepted Answer

After adding 150.0 mL of 0.500 M HNO3 to 300.0 mL of 0.569 M NH3, the solution is at the half-equivalence point. At this point, the pH is equal to the pKa of NH4+, which can be calculated using pKa = 14 - pKb. The pKb of NH3 is -log(1.8 x 10^-5) = 4.74, so pKa = 14 - 4.74 = 9.26. Therefore, the pH is approximately 9.36.

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

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