Consider the titration of 80.0 mL of 0.0200 M CH3NH2 (a weak base; Kb = 0.000440) with 0.100 M HNO3. Calculate the pH after the following volumes of titrant have been added:
(a) 0.0 mL
pH =
(b) 4.0 mL
pH =
(c) 8.0 mL
pH =
(d) 12.0 mL
pH =
(e) 16.0 mL
pH =
(f) 24.0 mL p
H =
Consider the titration of 80.0 mL of 0.0200 M CH3NH2 (a weak base; Kb = 0.000440)...
Consider the titration of 80.0 mL of 0.0200 M CH3NH2 (a weak base; Kb = 0.000440) with 0.100 M HNO3. Calculate the pH after the following volumes of titrant have been added: a) 0.0 ml b) 4.0 ml c) 8.0 ml d) 12.0 ml e) 16.0 ml f) 24.0 ml
38. Consider the titration of 80.0 mL of 0.0200 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HNO3. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = ________ (b) 4.0 mL pH = ___________ (c) 8.0 mL pH = ________ (d) 12.0 mL pH = __________ (e) 16.0 mL pH = __________ (f) 24.0 mL pH = _________
27. Consider the titration of 80.0 mL of 0.0200 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 4.0 mL pH = (c) 8.0 mL pH = (d) 12.0 mL pH = (e) 16.0 mL pH = (f) 30.4 mL pH =
Consider the titration of 80.0 mL of 0.0200 M NH3 (a weak base; Kb = 1.80e-05) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 4.0 mL pH = (c) 8.0 mL pH = (d) 12.0 mL pH = (e) 16.0 mL pH = (f) 20.8 mL pH =
Consider the titration of 80.0 mL of 0.0200 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 4.0 mL pH = (c) 8.0 mL pH = (d) 12.0 mL pH = (e) 16.0 mL pH = (f) 25.6 mL pH =
Consider the titration of 80.0 mL of 0.0200 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 4.0 mL pH = (c) 8.0 mL pH = (d) 12.0 mL pH = (e) 16.0 mL pH = (f) 30.4 mL pH = please help :) I'm super confused
Consider the titration of 40.0 mL of 0.0600 M CH3NH2 (a weak base; Kb = 0.000440) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (b) 6.0 mL pH = (C) 12.0 mL pH = pH = 11.8 * x (d) 18.0 ml (e) 24.0 ml (f) 43.2 ml pH = pH = pH =
Consider the titration of 20.0 mL of 0.0800 M H2NNH2 (a weak base; Kb = 1.30e-06) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = ? (b) 4.0 mL pH = ? (c) 8.0 mL pH = ? (d) 12.0 mL pH = ? (e) 16.0 mL pH = ? (f) 20.8 mL pH = ?
Consider the titration of 20.0 mL of 0.0800 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HIO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 4.0 mL pH = (c) 8.0 mL pH = (d) 12.0 mL pH = (e) 16.0 mL pH = (f) 22.4 mL pH=
Consider the titration of 20.0 mL of 0.0800 M HONH, (a weak base; Kb = 1.10e-08) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (C) 8.0 mL (b) 4.0 ml pH = pH = pH = (d) 12.0 mL (e) 16.0 ml (f) 20.8 ml pH = 1 pH = pH =