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e Edit View History Bookmarks Tools Help Masteringchemistry Assig- x httpsel/session masteringchemistry.com/myewaem/newlassignmentprobleml MW Chem 1412 Fall 2016 ssignment34 Chap 1LDH Equivale Problem 17.47 Problem 17.47 Calculate the pH at the equivalence point for sodium hydroxide (NaOH) titrating 0.210 Al solutions of each of the following bases with 0.210 M HBr Express your answer using two decimal places. pH Subrmit My Answers Give Up Part B hydroxylamine (NH2OH) Express your answer using two decimal places. Submit My Answers GveUD Part C Express your answer using two decimal places. e Shortcut My Answers Submit ve Folder Synchronization hics Properties. hics options BogideExplanation / Answer
The acid is a strong acid HBr with 0.210 M.
part A) it is with 0.210M NaOH , a strong base . So at the equivalence point the salt formed is NaCl which does not undergo hydrolysis, so the pH= 7 at equivalence point.
Part B
0.210M NH2OH , the Kb of this weak base is 9.1x 10-9
At the neutralization point
B + HBr ------> BH+ + Br-
0.210 0 0 0 initial concentrations
- 0.210 - - change
0 0 0.210 /2 0.210 /2 equilibrium concentrations(volume is doubled)
Thus at the equivalence [conjugate acid] or salt formed is 0.210/2
It is a salt formed from weak base and strong acidand therefore be acidic in aqueous solution due to cationic hydrolysis.
Such a solutions pH is given by
pH = 1/2 pKw -1/2pKb -1/2logC
= 7 - 1/2 ( 9-log 9.1) - 1/2 log (0.105)
= 1.9877
part c)0.210M aniline Kb of aniline is 4.3 x10-10
At the neutralization point
B + HBr ------> BH+ + Br-
0.210 0 0 0 initial concentrations
- 0.210 - - change
0 0 0.210 /2 0.210 /2 equilibrium concentrations(volume is doubled)
Thus at the equivalence [conjugate acid] or salt formed is 0.210/2
It is a salt formed from weak base and strong acidand therefore be acidic in aqueous solution due to cationic hydrolysis.
Such a solutions pH is given by
pH = 1/2 pKw -1/2pKb -1/2logC
= 7 - 1/2 (9.3665) -1/2 log 0.105
= 1.8274
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