Gradebook + Print Calculator -d Periodic Table Question I of 13 Incorrect Saplin
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Gradebook + Print Calculator -d Periodic Table Question I of 13 Incorrect Sapling Learning Map Caloulate the pH of a 0.343 M solution of ethylenediamine (H2NCH CH2NH2). The pK, values for the acidic form of ethylenediamine (H&NCH2CH2NH3;) are 6.848 (pKa) and 9,928 (pKa). Number pH 11.73 Calculate the concentration of each form of ethylenediamine in this solution at equilibrium Number HNCH,CH,NH,0338 Number [H,NCH,CH, NH0342995M Number [ |M HINCH,CH,NH0342995 incorrect. O PN-bus ) Give Up & View Solution # Try Again O Next Exit ExplanationExplanation / Answer
pKb1 = 14 - pKa2 = 14 - 9.928 = 4.072
Kb1 = 8.47 x 10^-5
pKb2 = 14 - 6.848 = 7.152
Kb2 = 7.5 x 10^-8
H2NCH2CH2NH2 = B
H2NCH2CH2NH2 + H2O --------------------->H2NCH2CH2NH3+ + OH-
0.343 -x x x ------------> at equilibrium
Kb1 = x^2 / 0.343-x
8.47 x 10^-5 = x^2 / 0.343-x
x^2 + 8.47 x 10^-5x -2.905 x 10^-5 = 0
x = 5.35 x 10^-3
[H2NCH2CH2NH2] = 0.343 -x
[H2NCH2CH2NH2] = 0.338 M
[H2NCH2CH2NH3+] = 5.35 x 10^-3 M
pOH = -log [OH-]
pOH = -log (5.35x 10^-3)
pOH = 2.27
pH + POH= 14
pH = 11.73
[+H3NCH2CH2NH3+] = pKb2 = 7.5 x 10^-8 M
note : all answers are bolded
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