QUESTION 12 Determine the volume in mL of 0.25 M HCl(aq) needed to reach the equivalence...
Determine the volume in mL of 0.45 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 35.2 mL of 0.31 M CH3CH2NH2(aq). Enter your answer with one decimal place. The Kb of ethylamine is 6.5 x 10-4.
Determine the volume in mL of 0.57 M HNO3(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 37.9 mL of 0.5 M CH3NH2(aq)(aq). The Kb of methylamine is 3.6 x 10-4. Enter your answer with two decimal places and no units. (how is the answer 16.62?)
What volume of 0.134 M H3PO4 (aq) in mL would be needed to reach the equivalence point in a titration with 20.00 mL of 0.367 M Ba(OH)2? Enter your answer with at least 3 sig figs. Assume that the units are mLs.
Determine the pH at the equivalence (stoichiometric) point in the titration of 33 mL of 0.22 M C2H5NH2(aq) with 0.16 M HCl(aq). The Kb of ethylamine is 6.5 x 10-4
Find the pH of the equivalence point and the volume (mL) of 0.175 M HCl needed to reach the equivalence point in the titration of 65.5 mL of 0.234 M NH3.
1)A 10.0 mL sample of 0.25 M NH3(aq) is titrated with 0.20 M HCl(aq) (adding HCl to NH3). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid.Kb of NH3 is 1.8 × 10−5.Henderson–Hasselbalch equation:Part a):1) After adding 10 mL of the HCl solution, the mixture is [ Select ] ["at", "before", "after"] the equivalence point on the titration curve.2) The pH of the solution after...
Question 6 1 pts A 10.0 mL sample of 0.25 M NH3(aq) is titrated with 0.20 M HCl(aq) (adding HCl to NH3). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid. Ko of NH3 is 1.8 x 10-5 Henderson-Hasselbalch equation: pH = pka + log og HCI NH, NH3- Parta): 1) After adding 10 mL of the HCl solution, the mixture is (Select] the equivalence...
Be sure to answer all parts. Find the pH and the volume (mL) of 0.487 M HNO3 needed to reach the equivalence point in the titration of 2.65 L of 0.0750 M pyridine (C5HzN, Kb = 1.7 x 10-9). Volume = mL HNO3 pH-
Determine the pH at the equivalence (stoichiometric) point in the titration of 48 mL of 0.28 M CH3NH2(aq) with 0.2 M HCl(aq). The Kb of methylamineis 3.6 x 10-4. The answer is 5.74. Please explain!
Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH3)3N(aq), with 0.1000 M HCl(aq) after 21 mL of the acid have been added. Kb of trimethylamine = 6.5 x 10-5.