https://padlet.com/valeria_aponte/b2sioqvlleq82z28 Valeria Aponte CHM1046-2233-5518 en-us 2023-04-16 19:46:40 UTC 2023-04-17 02:38:37 UTC hello@padlet.com https://padlet-uploads.storage.googleapis.com/2011244293/55bf20dc2ee9d0fcc7f72c22ac1a59c2/1E719D62_A268_4BC1_8687_6F649470007F.png valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556029582 A chemist is titrating 219.5 mL of a 0.1100 M piperidine (C5H10NH) solution with a 0.2500 M solution of HNO3. The pKb of piperidine is 2.89. Calculate the pH and the pH of the base solution after the chemist added 108.4mL of the HNO3 solution.

]]> 2023-04-16 22:22:59 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556029582 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556029985 Before diving in, take a breath, and begin from the basics. When calculating pH during the titration of a weak base with a strong acid, it can be separated into a solution stoichiometry problem and a pH problem. Let's Balance the equation to get our footing.

The chemical equation tells you it takes 1 mole of HNO3 to neutralize each mole of C5H10NH. Meaning the base is the limiting reaction and will run out first during the titration.]]> 2023-04-16 22:24:10 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556029985 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556030585 Use an ICE table to help the initial pH. Proceed by evaluating 2.89(pKb)=x^2/0.110-x. This equals to 0.5682 --> 
-log(0.5682)=0.24885. Initial pH=13.75]]> 2023-04-16 22:25:47 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556030585 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031097 Calculating the molarity of HNO3 and C5H10NH for equivalence:
C5H10NH-->0.1100 mol/L (0.2195 L)= 0.0241450 mol
HNO3--> 0.2500 mol/L (0.1084 L)=0.0271000 mol]]> 2023-04-16 22:27:18 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031097 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031438 0.2195L(0.1100M)=0.024145 will be the initial concentration
0.024145/0.2500=0.09658 L needed to reach the equivalence point 
Add Volumes-->0.09658 L+0.2195L =0.31608 L
0.024145/0.3168=0.07622]]> 2023-04-16 22:28:15 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031438 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031599 2023-04-16 22:28:41 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556031599 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556034713 Titration Problems can be intimidating when first approached, so it's very important to be patient and not overanalyze. They are intricate and detailed, almost like when one is baking from scratch. Hopefully, with this analogy in the steps below, problems like these will be a piece of cake!]]> 2023-04-16 22:37:01 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556034713 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556102182 Add the volumes together to find the total volume. Make sure to convert them. (mL-->L)]]> 2023-04-17 00:32:47 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556102182 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556133838 2023-04-17 01:03:36 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556133838 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556191943 2023-04-17 01:53:09 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556191943 valeria_aponte https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556217163 2023-04-17 02:13:08 UTC https://padlet.com/valeria_aponte/b2sioqvlleq82z28/wish/2556217163