Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.0 10-6) by 0.200 M

Question

Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.0 10-6) by 0.200 M HNO3. Calculate the pH of the resulting solution after 100.0 mL of HNO3 have been added. I know the pH is 1.301029996 I don't know how many significant figures to go to. Please give the answer in the correct number of sig figs. (Hint: I have tried 1, 1.3, and 1.30. None of them are correct.)

Explanation / Answer

moles H2NNH2 = 0.100 L x 0.100 M = 0.0100 moles HNO3 = 0.0200 L x 0.200 M = 0.00400 total volume = 0.120 L H2NNH2 + H+ >> H2NNH3+ moles H2NNH2 = 0.0100 - 0.00400 = 0.00600 concentration H2NNH2 = 0.00600 / 0.120 L = 0.0500 M concentration H2NNH3+ = 0.00400 / 0.120 = 0.0333 M pKb = 5.5 pOH = 5.5 + log 0.0333/ 0.0500 = 5.3 pH = 14 - 5.3 = 8.7 i think this is enough for u......

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---