Mast Chemistry: Chapter 17 Google Chrome https:// session masteringchemistry.com
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Mast Chemistry: Chapter 17 Google Chrome https:// session masteringchemistry.com ?offset next&assignment; ProblemID 65603235 M CBIRCKBICHLERCHEM 108SUMMER 2016 Chapter 17 Problem 17.47 Submit My Answers Give U Problem 17.47 Part B Calculate the pH at the equivalence point for titrating 0.180 M solutions of each of the following bases with 0.180 hydroxylamine (NH2OH) M HBr Express your answer using two decimal places pH Submit My Answers Give Up Part C aniline (C6H5NH2) Express your answer using two decimal places pH Submit My Answers Give Up Ask me anything Help Close Resources T previous l 10 of 11 l next Provide Feedback Continue 2:06 PM 7/22/2016Explanation / Answer
Calculate the ph at the equivalent point for titrating 0.180 M solution of each of the following bases with 0.180 MHbr
-hydroxylamine (NH2OH)
hydroxylamine (NH2OH) -
[H+] = [Ka*c]
pKas of NH3OH+ = 5.96
During titration
the weak base with a strong acid --------à a solution of the conjugate acid at half the concentration .
Plugging values in equation
= [(0.09 M) (10^-5.96)] = 1.16x10^-2
10^-5.96=0.00000109647
[H+] = 1.16x10^-2
=2 — log 1.16 = 2 — 0.06 = 1.94.
= pH = 1.94
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-aniline (C6H5NH2)
Ka for the conjugate acid of aniline
Ka = Kw/Kb for aniline
C6H5NH3+ = 4.60
During titration
the weak base with a strong acid --------à a solution of the conjugate acid at half the concentration
aniline (C6H5NH2) –
plugging values in equation
[H+] = [(0.09 M) (10^-4.60)] = 1.5x10^-3, or pH = 2.8.
[H+] = 1.5 x 10^-3,
pH = 2 -------------answer