Assume Kw = 1.0 x 1014 for each problem unless otherwise stated. (a) What are th
ID: 512399 • Letter: A
Question
Assume Kw = 1.0 x 1014 for each problem unless otherwise stated.
(a) What are the equilibrium concentrations of C6H5NH2, C6H5NH3+ , Cl, H+, and OH in an aqueous solution prepared by combining 50.0 mL of 0.0400 M C6H5NH3+Cl and 50.0 mL of 0.0100 M KOH? Kb = 7.4 x 1010 for C6H5NH2.
(b) Is the resultant solution in Part a above a buffer solution? Justify your choice.
(please give explanations for steps including the chemical reaction library and finding the Ka and Kb values, my teacher did not go over this slowly at all)
Explanation / Answer
Solution-
Moles of C6H5NH3+ present Initially= 50.0 mL * ( 1L/1000 L) * 0.0400 mol/L
= 2.00 * 10-3 mol
Moles of OH- present Initial= 50.0 mL * ( 1L/1000 L) * 0.0100 mol/L
= 5.00*10-4 mol
Preparing ICE table
C6H5NH3+ + OH- ---------> C6H5NH2 + H2O
I(mol) : 2.00 * 10-3 5.00*10-4 0
C(mol): -x -x +x
--------------------------------------------------------
E(mol) 2.00 * 10-3 - x 5.00 * 10-4 - x x
E(M)*0.05 2.00 * 10-3 - x 5.00 * 10-4 - x x
Kb of C6H5NH2 = 7.4*10-10
C6H5NH3+ + OH- ---------> C6H5NH2 + H2O , K = 1 / Kb = 1.4*109
K = (5.00 * 10-4 - x)( x) / (2.00 * 10-3 - x)(0.05)
x = 2.8*10-11 mol
Equilibrium concentrations are :[ C6H5NH2 ] = 5.6* 10^-10 M
[C6H5NH3+ ] = 0.04 M
[ OH- ] = 0.01 M
[H+] = 1.0*10^-14 / [ OH- ]
= 1.0*10-12 M
b) pH = -log[H+]
[H+] = 1.0*10^-14 / [ OH- ]
= 1.0*10-12 M
-log[1.0*10-12]
= 12
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