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i Safari File Edit View History Bookmarks Window Help 8 ?9% D' Tue 10:49:48 PM www-awh.aleks.com 2 My Academics- AU Access ALEKS Homework Link Here ALEKS Christian James-Learn ALEKS Ansley Davis-Learn nach3co2 molar mass-Google.. Chegg Study | Guided Solutions+ ACIDS AND BASES Ansley ktop Calculating the pH at equivalence of a titration A chemist titrates 250.0 mL of a 0.8863 M aniline (CoH,NH2) solution with 0.8979 M HBr solution at 25 °C. Calculate the pH at equivalence. The pK, of aniline is 4.87 Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added alo en Shot Ar Shot 6...2.24 PM fc 50 Shot 7.03 PM Explanation Check C 2018 McGraw Hill Education. All Rights Reserved. Terms of Use | Privacy 10Explanation / Answer
no of moles of C6H5NH2 = molarity * volume in L
= 0.8863*0.25 = 0.221575 moles
C6H5NH2 + HBr ---------------> C6H5NH3Br
1 mole of C6H5NH2 react with 1 mole of HBr
0.221575 moles of C6H5NH2 react with 0.221575 moles of HBr
no of moles of HBr = molarity * volume in L
0.221575 = 0.8979 * volume in L
volume in L = 0.221575/0.8979 = 0.2467L = 246.7ml
Total volume = 246.7 + 250 = 496.7ml = 0.4967L
molarity of C6H5NH3Br = no of moles/volume in L
= 0.221575/0.4967 = 0.446 M
C6H5NH3^+ (aq) + H2O --------------> C6H5NH2(aq) + H3O^+ (aq)
I 0.446 0 0
C -x +x +x
E 0.446-x +x +x
Pkb = 4.87
-logKb = 4.87
Kb = 10^-4.87 = 1.35*10^-5
Ka = Kw/Kb
= 1*10^-14/1.35*10^-5 = 7.4*10^-10
Ka = [C6H5NH2][H3O^+]/[C6H5NH3^+]
7.4*10^-10 = x*x/0.446-x
7.4*10^-10 *(0.446-x) = x^2
x = 1.8*10^-5
[H3O^+] = x = 1.8*10^-5M
PH = -log[H3O^+]
= -log1.8*10^-5
= 4.74
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